# C Program to find Roots of a Quadratic Equation

How to write a C Program to find Roots of a Quadratic Equation with an example?. A Quadratic Equation in C can have two roots, and they depend entirely upon the discriminant. The mathematical representation of a Quadratic Equation is ax²+bx+c = 0. If discriminant > 0 then Two Distinct Real Roots will exist for this equation

If discriminant = 0 then Two Equal and Real Roots will exists

and if discriminant < 0 then Two Distinct Complex Roots will exist

## C Program to find Roots of a Quadratic Equation Using Else If

This program allows the user to enter three values for a, b, and c. Next, this c program find roots of a quadratic equation using Else If Statement and sqrt function.

```/* C Program to find Roots of a Quadratic Equation */

#include <stdio.h>
#include<math.h>
int main()
{
float a, b, c;
float root1, root2, imaginary, discriminant;

printf("\n Please Enter values of a, b, c of Quadratic Equation :  ");
scanf("%f%f%f", &a, &b, &c);

discriminant = (b * b) - (4 * a *c);

if(discriminant > 0)
{
root1 = (-b + sqrt(discriminant) / (2 * a));
root2 = (-b - sqrt(discriminant) / (2 * a));
printf("\n Two Distinct Real Roots Exists: root1 = %.2f and root2 = %.2f", root1, root2);
}
else if(discriminant == 0)
{
root1 = root2 = -b / (2 * a);
printf("\n Two Equal and Real Roots Exists: root1 = %.2f and root2 = %.2f", root1, root2);
}
else if(discriminant < 0)
{
root1 = root2 = -b / (2 * a);
imaginary = sqrt(-discriminant) / (2 * a);
printf("\n Two Distinct Complex Roots Exists: root1 = %.2f+%.2f and root2 = %.2f-%.2f", root1, imaginary, root2, imaginary);
}

return 0;
}```

Within this program to find roots of quadratic equation example, User entered Values are 10 15 -25. It means a = 10, b = 15, c = -25 and the Quadratic equation is 10x²+15x-25 = 0

discriminant = (b * b) – (4 * a *c) = (15 * 15) – (4 * 10 *(-25))
discriminant = (225) – (- 1000) = 225 + 1000 = 1225

It means, discriminant > 0 so,
root1 = (-b + sqrt(discriminant) / (2 * a)) = (-15 + sqrt(1225) / (2 * 10))
root1 = (-15 + 35 / (20)) = -15 + 1.75 = -13.25

root2 = (-b – sqrt(discriminant) / (2 * a)) = (-15 – sqrt(1225) / (2 * 10))
root2 = (-15 – 35 / (20)) = -15 – 1.75 = -16.75

## C Program to Calculate Roots of a Quadratic Equation Using Switch Case

This program to find roots of a quadratic equation is the same as above, but this time we are using the Switch case in C Programming.

```/* C Program to find Roots of a Quadratic Equation using Switch Case */
#include <stdio.h>
#include<math.h>
int main()
{
float a, b, c;
float root1, root2, imaginary, discriminant;

printf("\n Please Enter values of a, b, c of Quadratic Equation :  ");
scanf("%f%f%f", &a, &b, &c);

discriminant = (b * b) - (4 * a *c);

switch(discriminant > 0)
{
case 1:
root1 = (-b + sqrt(discriminant) / (2 * a));
root2 = (-b - sqrt(discriminant) / (2 * a));
printf("\n Two Distinct Real Roots Exists: root1 = %.2f and root2 = %.2f", root1, root2);
break;
case 0:
switch(discriminant < 0)
{
case 1: //True
root1 = root2 = -b / (2 * a);
imaginary = sqrt(-discriminant) / (2 * a);
printf("\n Two Distinct Complex Roots Exists: root1 = %.2f+%.2f and root2 = %.2f-%.2f", root1, imaginary, root2, imaginary);
break;
case 0: // False (Discriminant = 0)
root1 = root2 = -b / (2 * a);
printf("\n Two Equal and Real Roots Exists: root1 = %.2f and root2 = %.2f", root1, root2);
break;
}
}

return 0;
}```

C Roots of a Quadratic equation Output

`````` Please Enter values of a, b, c of Quadratic Equation :  2 3 5

Two Distinct Complex Roots Exists: root1 = -0.75+1.39 and root2 = -0.75-1.39``````

Within this C Program to find Roots of a Quadratic Equation example, User entered Values are 2 3 5. It means a = 2, b = 3, c = 5 and the Quadratic equation is 2x²+3x+5 = 0

discriminant = (b * b) – (4 * a *c)  = (3 * 3) – (4 * 2 * 5)
discriminant = (9) – (40) = -31

It means, discriminant < 0 so,
root1 = root2 = -b / (2 * a)
root1 = root2 = -3 / (2 * 2) = -0.75

imaginary = sqrt(-discriminant) / (2 * a)
imaginary = sqrt(- -31) / (2 * 2) =  5.567 / 4 = 1.3919

root1 = root1 + imaginary
root1 = -0.75+1.3919

root2 = root2 – imaginary
root2 = -0.75-1.3919