In this C program, we will find roots of a quadratic equation. A quadratic equation is a second order equation having a single variable. Any quadratic equation can be represented as ax^{2} + bx + c = 0, where a, b and c are constants( a can't be 0) and x is unknown variable.

##### For Example

2x^{2}+ 5x + 3 = 0 is a quadratic equation where a, b and c are 2, 5 and 3 respectively.

To calculate the roots of quadratic equation we can use below formula. There are two solutions of a quadratic equation.

**x = (-2a + sqrt(D))/2
x = (-2a - sqrt(D))/2 **

where, D is Discriminant which is calculated as (b

^{2}- 4ac), it differentiate the nature of the roots of quadratic equation.

Discriminant(D) value | Description |
---|---|

D < 0 | We will get two complex roots. |

D = 0 | We will get two equal roots. |

D > 0 | We will get two real numbers. |

## C program to find all roots of a quadratic equation

#include <stdio.h> #include <math.h> int main() { float a, b, c, determinant, root1, root2, real, imag; printf("Enter coefficients a, b and c\n"); scanf("%f%f%f", &a, &b, &c); /* Calculate determinant */ determinant = b*b - 4*a*c; if(determinant >= 0) { root1= (-b + sqrt(determinant))/(2 * a); root2= (-b - sqrt(determinant))/(2 * a); printf("Roots of %.2fx^2 + %.2fx + %.2f = 0 are \n%.2f and %.2f", a, b, c, root1, root2); } else { real= -b/(2*a); imag = sqrt(-determinant)/(2 * a); printf("Roots of %.2fx^2 + %.2fx + %.2f = 0 are \n%.2f+%.2fi and %.2f-%.2fi",a,b,c,real,imag,real,imag); } return 0; }Output

Enter coefficients a, b and c 1 1 1 Roots of 1.00x^2 + 1.00x + 1.00 = 0 are -0.50+0.87i and -0.50-0.87i Enter coefficients a, b and c 3 7 2 Roots of 3.00x^2 + 7.00x + 2.00 = 0 are -0.33 and -2.00

**Related Topics**