C program to Find all Roots of a Quadratic equation

This program accepts coefficients of a quadratic equation from the user and displays the roots (both real and complex roots depending upon the determinant).
Quadratic equation graph

To understand this example, you should have the knowledge of following C programming topics:

The standard form of a quadratic equation is:

ax2 + bx + c = 0, where
a, b and c are real numbers and
a ≠ 0

The term b2-4ac is known as the determinant of a quadratic equation. The determinant tells the nature of the roots.

  • If determinant is greater than 0, the roots are real and different.
  • If determinant is equal to 0, the roots are real and equal.
  • If determinant is less than 0, the roots are complex and different.

Calculation of roots of a quadratic equation

Example: Program to Find Roots of a Quadratic Equation

#include <stdio.h>
#include <math.h>

int main()
    double a, b, c, determinant, root1,root2, realPart, imaginaryPart;

    printf("Enter coefficients a, b and c: ");
    scanf("%lf %lf %lf",&a, &b, &c);

    determinant = b*b-4*a*c;

    // condition for real and different roots
    if (determinant > 0)
    // sqrt() function returns square root
        root1 = (-b+sqrt(determinant))/(2*a);
        root2 = (-b-sqrt(determinant))/(2*a);

        printf("root1 = %.2lf and root2 = %.2lf",root1 , root2);

    //condition for real and equal roots
    else if (determinant == 0)
        root1 = root2 = -b/(2*a);

        printf("root1 = root2 = %.2lf;", root1);

    // if roots are not real 
        realPart = -b/(2*a);
        imaginaryPart = sqrt(-determinant)/(2*a);
        printf("root1 = %.2lf+%.2lfi and root2 = %.2f-%.2fi", realPart, imaginaryPart, realPart, imaginaryPart);

    return 0;


Enter coefficients a, b and c: 2.3
Roots are: -0.87+1.30i and -0.87-1.30i

In this program, library function sqrt() is used to find the square root of a number. To learn more, visit:  sqrt() function.