How to calculate square/cubic root by hand?

Today, technologies make everything easy and fast for us. Sometimes, when we see the ancient marvels of engineering, we wonder: “How they designed this without computers”? And the answer is: “with knowledge, inspiration and hard work”.

However, being almost irreplaceable, computers makes us more and more adjective. Our brains become lazy and we are losing some knowledge and skills. Such is the manual calculation of root. It is so easy and elegant, so I will share it here for the sake of the good old math science.

Square root

Lets calculate x = √a first.

By squaring both sides and moving a to the left side, we obtain the following equation:

x2a = 0

This is an equation of type f(x) = 0 and we can solve numerically it using the Newton’s method:

1. Start with initial guess: x0. It is good to be close to the solution, so you can use the nearest exact root.
2. Calculate the next approximation of the root by the equation: x1 = x0f(x0)/f′(x0)
3. Continue the iterations by using the previous result to estimate a new one:
xn+1 = xnf(xn)/f′(xn).
4. Stop when you reach the desired precision.

In the above formulas, f(x) is the function and f′(x) is the first derivative. In our case:

f(x) = x2a

f′(x) = 2x

If we substitute the above equations into the Newton’s iterative formula, we get:

xn+1 = xn − (xn2a)/2xn

xn+1 = (xn + a/xn)/2

Now, lets use it to calculate √a = ?, for a = 5.

We will start with an initial guess of x0 =√4 = 2.

1st iteration: x1 = (x0 + a/x0)/2 = (2 + 5/2)/2 = 2.25

2nd iteration: x2 = (x1 + a/x1)/2 = (2.25 + 5/2.25)/2 = 2.2361

3rd iteration: x3 = (x2 + a/x2)/2= (2.2361 + 5/2.2361)/2 = 2.2361

After the second iteration, we have 5 correct significant digits, which is quite sufficient for most cases.

Cubic root

We will use the same approach, as follows:

f(x) = x3a

f′(x) = 3x2

xn+1 = xn − (xn3a)/3xn2

xn+1 = (2xn + a/xn2)/3

Lets see how it works for 3√7. Our initial guess is: x0 = 3√8 = 2.

1st iteration: x1 = (2x0 + a/x02)/3 = (2·2 + 7/22)/3 = 1.9167

2nd iteration: x2 = (2x1 + a/x12)/3 = (2·1.9167 + 7/1.91672)/3 = 1.9129

3rd iteration: x3 = (2x2 + a/x22)/3 = (2·1.9129 + 7/1.91292)/3 = 1.9129

Again, we managed to find it after two iterations. We can also use Calcpad to create a tiny program for that:

It is interesting that just after 4 iterations, the Newton’s method finds 16 correct significant digits which is the limit precision for most computers. We also calculated the root by three different methods which give the same result.

How to plot the Julia Set

In the previous post we discussed the Mandelbrot set. There is another beautiful algebraic fractal, called Julia set 𝓙. It is named after the French mathematician Gaston Julia, who has discovered. It is obtained by the same formula as the Mandelbrot Set:

zn+1 = zn2 + c

The difference is that z0 = x + yi is the current point (x, y) in the complex plane and c is a fixed complex number, while for the Mandelbrot set z0 = 0 and c = x + yi.

Both sets are closely related. There are a lot of different Julia sets, corresponding to different values of c. Their shapes and properties depend on where the point is located, related to the Mandelbrot set 𝓜. Inside 𝓜, the corresponding Julia set 𝓙 is connected. Outside 𝓜, it falls apart into a set of isolated points, called Fatou set 𝓕. Further away, it turns into Fatou dust – infinite number of points with zero area, similarly to the Cantor dust.

For simple plotting, you can use the following code:

```"Julia set
'Define the function
JuliaSet(z; c) = \$Repeat{z = z^2 + c @ i = 1 : n}
#hide
'Set the plotting parameters
PlotStep = 1','PlotWidth = 600','PlotHeight = 400
#show
'Plot for'c = -0.4 + 0.59i'and'n = 200'iterations
\$Map{abs(JuliaSet(x + 1i*y; c))) @ x = -1.5 : 1.5 & y = -1 : 1}
```

When you run it inside Calcpad, you will get the following result:

To obtain better and colorful images, we will use different approach this time. We will modify the JuliaSet function in a way to return directly the number of iterations for which the result overpasses 2. The source code is provided bellow:

```"Julia set
J(z; c; n) = abs(\$Repeat{z = if(abs(z) < 2; z^2 + c; 4) @ i = 1 : n})
JuliaSet(z; c) = \$Find{J(z; c; x) - 2 @ x = 1 : n}
#hide
PlotStep = 1','PlotWidth = 600','PlotHeight = 400','Precision = 10^-6
#show
'Plot for'c = -0.4 + 0.6i'and'n = 100'iterations
\$Map{JuliaSet(x + 1i*y; c)) @ x = -1.5 : 1.5 & y = -1 : 1}
```

It runs a little bit slower, but the result is worth the wait:

You can download Calcpad for free and experiment by yourself with different values of c. For example, the lightning image bellow is a special case of Julia set for c = i and is called “dendrite” fractal.

Plotting the Mandelbrot Set

Calcpad has an interesting and undocumented feature, that I am going to reveal in this post. You can use it to quickly plot the Mandelbrot set. This is a set of complex numbers c, for which the iterative equation zn+1 = zn2 + c does not go to infinity. The most beautiful part is that it shows fractal behavior at its boundaries. If you plot it with the appropriate colors, you can create stunning images and animations. I really like those on the Maths Town’s YouTube channel:

Usually, such videos are created by complicated software, specially developed for that purpose. However, you can also plot the Mandelbrot set with Calcpad and a few lines of code. There are two ways to do that. The first one is to use the Calcpad capabilities for complex arithmetic, as follows:

```"Mandelbrot set
'Define the function
MandelbrotSet(z; c) = \$Repeat{z = z^2 + c @ i = 1 : n}
#hide
'Set the plotting parameters
PlotStep = 1','PlotWidth = 500','PlotHeight = 500
#show
'Plot for'n = 50'iterations
\$Map{abs(MandelbrotSet(0; x + 1i*y)) @ x = -1.5 : 0.5 & y = -1 : 1}
```

Open Calcpad, switch to “complex”mode, paste the above code inside, and run the calculations. You will get the following result:

It is not too bad, but still not very impressive. At least, the exterior is missing any colors. Then, how we can do the coloring? Theoretically, it is satisfied for all numbers in the set that |zn| < 2 for any n. Outside, the process diverges and the modulus goes to infinity, as n increases. You can use that in the following way:
1. For each point, calculate the number of iterations n, where |zn| gets larger than 2.
2. Map these values to colors.
3. Plot the colors at the respective coordinates.

If you play with the number of iterations n in the above example, you will see that the plot changes. The shaded area is where |zn| converges for the respective n.

If you combine the above plots and apply some colors, you can get really beautiful artistic image like the one bellow:

The only problem with this method is that it is slow and elaborate. That is why, I included a special function Mandelbrot(x; y) in Calcpad. It calculates the number of iterations, needed for the modulus to overpass 2. You can plot this function directly with the preferred color scale:

To do that, I used even shorter piece of code:

```"Mandelbrot set
#hide
'Set the plotting parameters
PlotStep = 1','PlotWidth = 500','PlotHeight = 500
#show
'Plot for'n = 50'iterations
\$Map{Mandelbrot(x; y) @ x = -1.5 : 0.5 & y = -1 : 1}
'Plot for'n = 70'iterations
\$Map{Mandelbrot(x; y) @ x = -0.59 : -0.58 & y = 0.55 : 0.56}
```

Here you can find the C# code of the Mandelbrot function that works inside Calcpad:

```        private static readonly double Log2Inv = 1 / Math.Log(2);
//Calculates the number of iterations for which z = z^2 + c satisfies |z| > 2
//Then transforms it to a smooth log scale and returns the result
protected static double MandelbrotSet(double x, double y)
{
//Checks if the point is inside the set and directly returns NaN
if (x > -1.25 && x < 0.375)
{
if (x < -0.75)
{
if (y > -0.25 && y < 0.25)
{
double x1 = x + 1,
y2 = y * y,
x2 = x1 * x1;
if (x2 + y2 <= 0.0625)
return double.NaN;
}
}
else if (y > -0.65 && y < 0.65)
{
double x1 = x - 0.25,
y2 = y * y,
q = x1 * x1 + y2;
if (q * (q + x1) <= 0.25 * y2)
return double.NaN;
}
}
//For all other points performs detailed calculations
double re = x, im = y;
for (int i = 1; i <= 1000; ++i)
{
double reSq = re * re,
imSq = im * im,
sumSq = reSq + imSq;
//To avoid the sqrt function, the check |z| > 2 is replaced by z^2 > 4
if (sumSq > 4)
{
var logZn = Math.Log(sumSq) / 2;
var nu = Math.Log(logZn * Log2Inv) * Log2Inv;
return (1.01 - Math.Pow(i - nu, 0.001)) * 1000;
}
//Calculates z = z^2 + c
im = 2 * re * im + y;
re = reSq - imSq + x;
}
return double.NaN;
}
}
```

The complete source code is available on GitHub.

Even more 3D effects

In the last version I improved the 3D plot shading and added some nice 3D effects. Bellow is the plot of the function

f(x; y) = cos(x) + cos(y) + cos(x·y/3)/2 + cos(√(x2 + y2)·2)/2

with the “Rainbow” color scale and the “Shadows” option on:

What is new is the specular light effect (the little shiny reflections). I also replaced the Gouraud shading procedure with the Phong one. It improved the image quality  without any noticeable impact on the speed (surprisingly). Since the light is distant and the view is isometric, I use the Blinn–Phong shading model. Instead of calculating the reflection vector for each vertex, it calculates the halfway vector between the viewer and light-source. It is performed once for the whole model, and that is why it is faster.

In general, there is additional overhead for generating the Html output, converting the image to base64 and rendering the Html into the “Output” window. However, as long as it takes much less than a second for all of it, the speed is not of a big concern.

Most people consider math boring, but sometimes it can be really beautiful. Recently I found some unusual application of Calcpad – to draw nice pictures using math formulas. In general, you can plot functions of two variables using the \$Map command:

`\$Map{f(x;y) @ x = a : b & y = c : d}`

I notices that you can get nice effects using the following types of functions:

`f(x;y) = abs(cos(p1(x;y)) + cos(p2(x;y)) ...)`

It creates families of intersecting curves that from canyons with hills in between. If you select “none” for the color scale, it creates nice and smooth pictures with shadows only. Bellow are some examples:

f(x;y) = abs(cos(x^2/10) + cos(y^2/10))

f(x;y) = abs(cos(x + y/3) + cos(x*y/7))

f(x;y) = abs(cos(x – sin(y)) + cos(y – sin(x)))

f(x;y) = abs(cos(r(x;y)*e^(ξ*r(x;y)))*e^(-1.5*ξ*r(x;y)))

f(x;y) = abs(cos(r(x;y)) + cos(y))

f(x;y) = abs(cos(x) + cos((y/4)^2))

f(x;y) = abs(cos(2*r(x;y)) + cos(x) + cos(y))

f(x;y) = abs(cos(2*r(cos(x/2);cos(y/2))))

f(x;y) = abs(r(cos(x/2);cos(y/2)))

In the above equations, r(x;y) = sqr(x^2 + y^2).

Calcpad version 3.0 already supports some non-metric units, that are commonly used in science, technology and engineering. A list of currently supported units is available in Units section of the User manual. More units can be easily appended in the next version. We will appreciate your suggestions very much.

Calcpad supports both Imperal and USCS system. Some of the units have slightly different values although their names are the same. You can select which system to use by switching the respective option in the program window (●Imperial/○US).

You can convert between metric and non-metric units. That is possible because both systems exist simultaneously inside the Calcpad core engine. You can also mix them in one and the same expression. Open Calcpad on your desktop or go to the online calculator and try the following examples:

2ft + 3in|cm

kg/lb

You will get the following answer in the output window:

2ft + 3in = 68.58cm

kg/lb = 2.204623

Converting temperatures is a little bit more tricky because the origins of the different temperature scales are not the same. For example: 0°C = 32°F. Because of that, it makes a difference whether you convert between absolute values or temperature differences. In the next post, we will discuss more about this and how this problem is solved in Calcpad.

How units work

In this post, we will take a quick look inside Calcpad and see how units actually work. But first, let’s make a brief overview of the theory basis.

Theory basis

There are only seven base units in the whole universe. They correspond to the seven physical dimensions:

• mass – kilogram (kg)
• length – meter (m)
• time – second (s)
• electric current – ampere (A)
• temperature – Kelvin (K)
• amount of substance – mole (mol)
• luminous intensity – candela (cd)

All other units are derived from the above ones, using the respective laws of physics. For example, forces are measured in Newtons (N). By definition, force is mass × acceleration, so 1 N = 1 kg·m/s2. Such units are called “composite”. They can be always decomposed to their constituent units.

Other method to derive units is by adding prefixes. I this way, we can obtain multiples of both base and composite units. For example,  kilonewtons – kN = 103·N, meganewtons – MN = kN = 106·N and so on. All possible prefixes are listed in the following tables:

Name deca hecto kilo mega giga tera peta exa zetta yotta
Symbol da h k M G T P E Z Y
Factor 101 102 103 106 109 1012 1015 1018 1021 1024
Name deci centi milli micro nano pico femto atto zepto yocto
Symbol d c m μ n p f a z y
Factor 10−1 10−2 10−3 10−6 10−9 10−12 10−15 10−18 10−21 10−24

However, not all prefixes are commonly used for all units, and that is why they are not included in Calcpad. Full list of the available units is provided in the user manual:

Internal implementation

Any software implementation must allow the units to be compared, converted and scaled. For that purpose, we need to store the information for how units are composed by the base ones. We also need the corresponding scale factors, in order to allow for multiples. Then, we have to define all possible operations like comparison, addition, subtraction, multiplication, division and exponentiation.

To store all this information, we need a special type of data called “structure”. It can contain multiple values of different types in a single variable. In the case of units, we will include a name and two vectors. The first one will store the exponents for all seven physical dimensions (mass, length, time, current, temp, substance, luminosity) and the second one – the respective scaling factors:

`Unit = {name: "name", exponents: [p1, p2, p3, p4, p5, p6, p7], factors: [f1, f2, f3, f4, f5, f6, f7]}`

A base unit can include only one value of “1” for the respective dimension. All the rest must be exactly zero “0”. Composite units can have several nonzero values for the corresponding dimensions. Bellow are some examples of unit definitions:

Base units:

```Kilogram = {name: "kg", exponents: [1, 0, 0, 0, 0, 0, 0], factors: [103, 1, 1, 1, 1, 1, 1]} Meter = {name: "m", exponents: [0, 1, 0, 0, 0, 0, 0], factors: [1, 1, 1, 1, 1, 1, 1]} Second = {name: "s", exponents: [0, 0, 1, 0, 0, 0, 0], factors: [1, 1, 1, 1, 1, 1, 1]}```

Composite units:

```Newton = {name: "N", exponents: [1, 1, -2, 0, 0, 0, 0], factors: [1, 1, 1, 1, 1, 1, 1]} Pascal = {name: "Pa", exponents: [1, -1, -2, 0, 0, 0, 0], factors: [1, 1, 1, 1, 1, 1, 1]}```

Multiples:

```Ton = {name: "t", exponents: [1, 0, 0, 0, 0, 0, 0], factors: [106, 1, 1, 1, 1, 1, 1]} Millimeter = {name: "mm", exponents: [0, 1, 0, 0, 0, 0, 0], factors: [1, 10-3, 1, 1, 1, 1, 1]} Minute = {name: "min", exponents: [0, 0, 1, 0, 0, 0, 0], factors: [1, 1, 60, 1, 1, 1, 1]} Kilonewton = {name: "kN", exponents: [1, 1, -2, 0, 0, 0, 0], factors: [103, 1, 1, 1, 1, 1, 1]} Megapascal = {name: "MPa", exponents: [1, -1, -2, 0, 0, 0, 0], factors: [106, 1, 1, 1, 1, 1, 1]}```

Units consistency

You can compare, add or subtract only values, which units are consistent. For example, “kg + m” does not make any sense, because both operands represent different physical dimensions. The expression “m2 + m3” does not make sense as well, because you are trying to add area to volume. However, “m + mm” is possible, because both represent length and we can convert them to each other. So, units are consistent only if they are of the same dimensions and with the same exponents. What we need to do for our structure, is to check if the exponents vectors match exactly.

Even if consistent, units are not necessarily equal, because they can have different scaling factors. So, we still cannot operate directly with the respective values. First, we have to convert all values to the same units. For example:

`3 dm + 2 cm = 3 dm + 0.2 dm = (3 + 0.2) dm = 3.2 dm`

Subtraction and comparison are performed in the same way as the addition. The following algorithm can be used:

1. Check if the units are consistent (compare the exponent vectors). If not, report an error. Otherwise, continue to step 2.
2. Compare the scale factors. If different, compute the conversion factor for the second unit and use it to multiply the second value.
3. Perform the operation and return the result with the units of the first value.

Multiplication, division and exponentiation

Unlike the previous ones, these operations do not require the units to be consistent. Actually, this is the way we use to derive new units by combining the existing ones. In the software, these operations are performed by manipulation of the respective exponents as follows:

1. Multiplication is performed by addition of the exponents:
`m·m2 = m(1 + 2) = m3`
2. Division is equivalent to exponent subtraction:
`N/m2 = kg·m·s-2/m2 = kg·m(1 - 2)·s-2 = kg·m-1·s-2 = Pa`
3. Exponentiation is performed by multiplication of the old and new exponent:
`(m2)3 = m2·3 = m6`

The following algorithm is applied:

1. Check if there are different scale factors and compute the conversion factor.
2. Perform the required operation (multiplication, division …) with the units and obtain the resulting units.
3. Perform the operation with the values, apply the conversion factor and return the result in the calculated units.

Target units

If you do not specify any target units, the result will be decomposed to the base units. For example:

`100N·10m = 1000 kg·m2/s2`

That is because the program cannot understand the physical meaning of your formulas. Sometimes, the result can be ambiguous and the program will not know what to do. For the above expression:

• If it is bending moment or torque, the result will be in Nm.
• If it is work or energy, the result will be in Joules.

In such cases, you have to specify the target units and if they are compatible, the result will be converted automatically. Always write the target units at the end of the expression, separated by a vertical bar “|”:

`500N·20m|kJ = 10kJ`

`2000N·5m|kNm = 10kNm`

Calculating with units

Unit conversion is one of the most tedious jobs when doing calculations. Although not very difficult, it is repetitive and time consuming.

For example, lets calculate the stress in a steel beam. Assume that span length is in meters, section dimensions are in mm, geometrical properties are in cm2 (cm3, cm4), section forces are in kN (kNm) and stress is in MPa. That requires you to include a lot of conversion factors in your formulas. So, it would be very helpful if you could write the units next to the respective values and let the software do all the conversions for you.

The most recent version Calcpad Pro 3 supports physical units in calculations. For now, only metric units (SI compatible) are included, but more units will be added in the next versions.

You can attach units to every value in an expression and obtain the result in the specified target units. The target units are added at the end, separated by a vertical bar “|”.  , Open the online calculator and try the following expression by yourself:

`2m + 50 cm + 200 mm|dm`

Since the target units are “dm“, the answer is 27dm (not just adding 20 + 50 + 200 = 252). That is because all values are converted automatically to the same units before the addition: 20dm + 5dm + 2dm = 27dm. If you do not specify any target units, the units of the first operand will be used.

If you want to do simple unit conversion, just write the value, followed by the source units, a vertical bar and the target units:

`25m/s|km/h`

The result is 90km/h, which is correct.

You can also attach units to variables and functions. If you have units in an expression that defines a variable, the result will be stored in the variable, together with the calculated units. If you specify target units at the end, they will be used instead. From this point further, the variable will be substituted always with the internally stored units. You can try the following examples:

Input Output
Example 1: Speed
```'Distance -'S = 50m
'Time -'t = 2s
'Speed -'V = S/t|km/h```
Distance – S = 50m
Time – t = 2s
Speed – V = S/t = 50m/2s = 90km/h
Example 2: Force
```'Mass - 'M = 500t
'Acceleration - 'a = 4m/s^2
'Force -'F = M*a|kN```
Mass – M = 500t
Acceleration – a = 4m/s2
Force – F = M·a = 500t·4m/s2 = 2000kN
Example 3: Stress in column
```'Column load -'F = 2000kN
'Section size -'b = 500mm
'Section area -'A = b^2|cm^2
'Stress -'σ = F/A|MPa```
Column load – F = 2000kN
Section size – b = 500mm
Area – A = b2 = (500mm)2 = 2500cm2
Stress – σ = F/A = 2000kN/2500cm2 = 8MPa

If you specify target units after a function definition, they will be attached permanently to this function. The result will be converted to them whenever the function is used further. However, you have to be careful to provide the proper units for the arguments. Otherwise, you can get an error.

Although Excel is still the most popular platform for engineering spreadsheets, it lacks such advantages as automated units. I think that math software is more appropriate for engineers in general, but this will be a topic for future discussions.

Read the next post to see how do units actually work. We will lift the cover and have a quick look inside the Calcpad engine, together.

In the previous post we created a simple program to solve a quadratic equation. We used the well known method from school:

```"Quadratic equation:
'a*x^2 + b*x + c = 0
a = 3 ', ' b = 4 ', ' c = -5
'Discriminant - ' D = b^2 - 4*a*c
x_1 = (-b - sqr(D))/(2*a)
x_2 = (-b + sqr(D))/(2*a)
```

In this post we will make our program more professional and add some advanced features. Currently, it is fine for personal use, but if we want to give it to other users, we need to work a little bit harder.

For example, let’s try to enter `a = 0` and see what happens. In fact, it will turn into a linear equation, which solution is `x = -c/b`. However, our program will end up with error due to division by zero. Calcpad will show `-∞, +∞` or `undefined`, if we have `0/0` as it will happen for `x_2`. It is similar for the case of a negative discriminant `D < 0` (unless we use complex numbers).

This is not very good. Every professional program must work correctly for any possible input data. It should either provide a sensible result or a comprehensible message to the user. But how to do this with Calcpad?

The good news is that Calcpad supports boolean expressions and program flow control as follows:

```#if
<code>
#else if
<code>
...
#else
<code>
#end if
```

This is a very powerful feature. It allows you to split the solution entirely, depending on some intermediate values. It is not just about using different formulas. You can even display different text and images into the output. It is very difficult to do this with Excel.

Dots “`...`” mean that you can add as many conditions as you need (do not place the dots into the code) Instead of `<condition>` you can put any expression that compares values, e.g. `a ≡ 0` or `D < 0`. So, lets add some logic into our program as follows:

```"Quadratic equation:
'a*x^2 + b*x + c = 0
a = 3 ', ' b = 4 ', ' c = -5
#if a ≡ 0
'The coefficient a is equal to zero. The equation is linear:
x = -c/b
#else
'Discriminant - ' D = b^2 - 4*a*c
#if D ≥ 0
'The discriminant is greater or equal to zero. Real roots exist.
q = -0.5*(b + sign(b)*sqr(D))
x_1 = q/a
x_2 = c/q
#else
'The discriminant is less than zero. There are no real roots.
#end if
#end if
```

Now, the program will display the correct answer in any case. You can copy the above script into Calcpad and try for `a = 0` or `b = 5`. We also made another improvement: The classical formula `x1,2 = (-b ± sqr(D))/(2·a)` is numerically unstable. The answer will be inaccurate due to the approximate representation of floating point numbers by the computer. The new formula solves this problem.

So far, so good. We managed to improve our program, but it is still not very convenient for other people to use it. When the program gets larger and more complicated, it becomes difficult for users to work directly with the source code and find where to change the input values. There is also a risk to accidentally damage some formulas and produce incorrect results.

That is why, Calcpad provides a quick way to generate input forms directly from the source code. All you need is to put question marks `?` wherever you want to enter values. In our program, we need to change the third line as follows:

`a = ? ', ' b = ? ', ' c = ?`

Then press the “form” button (left to the orange arrow) and you will get the input form:

But how we made the solution details to disappear during the input? In Calcpad, you can select which part of the output will be visible at each stage by using special keywords:

• `#hide` – always hide the content after this place ;
• `#pre`   – show the content only in input mode (before calculations);
• `#post` – show the content only in output mode (after calculations);
• `#show` – always show the remaining content (cancel);

If you want to hide all calculation details during the input, just add `#post` after the third line. There is also another method for hiding. Since all the output is finally formatted as Html, you can include Html tags and CSS in Calcpad comments. You can use this to enclose some contents in Html comments and it will hide as well:

`'<!--''-->`

Do not use this to hide sensitive information because it can be seen in the Html source. For that purpose, we strongly recommend to use `#hide`. It entirely skips the content in the output Html.

So, our first program is ready. It is pretty much like the one on the official Calcpad website:

The only additional thing is the function graph, but I am going to write an whole new post about graphing. Bellow is the final source code for the quadratic equation program. You can copy it an use it freely:

```"Quadratic equation:
'a·x<sup>2</sup> + b·x + c = 0
a = ? ', ' b = ? ', ' c = ?
#post
#if a ≡ 0
'The coefficient a is equal to zero. The equation is linear:
x = -c/b
#else
'Discriminant - ' D = b^2 - 4*a*c
#if D ≥ 0
'The discriminant is greater or equal to zero. Real roots exist.
q = -0.5*(b + sign(b)*sqr(D))
x_1 = q/a
x_2 = c/q
#else
'The discriminant is less than zero. There are no real roots.
#end if
#end if
```

My first program – quadratic equation

Lets write our first program with Calcpad. Most programming languages usually start with “Hello world!”, but for Calcpad, there is nothing much to do, except to write the text in quotes:

```"Hello world from Calcpad!"
```

Then press Enter and the text will appear in the output.

So, let’s do something more useful and solve a real problem. The quadratic equation is simple enough to start with.

All you need to do is to enter the formulas and the text in the “Script” box on the left. Then press Enter or the orange arrow and you will get the results in the “Output” box on the right:

Bellow, you can copy the source code and try it by yourself:

```"Quadratic equation:
'a*x^2 + b*x + c = 0
a = 3 ', ' b = 4 ', ' c = -5
'Discriminant - ' D = b^2 - 4*a*c
x_1 = (-b - sqr(D))/(2*a)
x_2 = (-b + Sqr(D))/(2*a)
```

All comments must be enclosed in quotes: double quotes for “titles” and single quotes for ‘simple text’. In this example, `"Quadratic equation` is a title and `'Discriminant - '` is text. The closing quote can be omitted, if there is nothing further on the line.

Everything outside the quotes is assumed to be math expressions.  Normally, each expression should be on a new line. If they are several expressions on the same line, they must be separated by comments, e.g.

`a = 3 ', ' b = 4 ', ' c = -5`

Expressions are defined by using the commonly accepted mathematical notation: numbers, variables, operators “`^ / \ % * - + ≡ ≠ ≤ ≥ =`” and standard mathematical functions. A short description of the available syntax is initially displayed in the output box. For more information, please check the user manual: