Please take screenshots of the program **Section 3.2****3.1 (Algebra: solve quadratic equations)**The two roots of a quadratic equation \( ax^2 + bx + c = 0 \) can be obtained using the following formula:\[r_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \text{and} \quad r_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\]\( b^2 - 4ac \) is called the discriminant of the quadratic equation. If it is positive, the equation has two real roots. If it is zero, the equation has one root. If it is negative, the equation has no real roots.Write a program that prompts the user to enter values for \( a \), \( b \), and \( c \) and displays the result based on the discriminant. If the discriminant is positive, display two roots. If the discriminant is 0, display one root. Otherwise, display “The equation has no real roots”.Note that you can use Math.pow(x, 0.5) to compute \( \sqrt{x} \). Here are some sample runs.

Following is the code for finding the roots of the equation under the conditions mentioned#include <stdio.h>#include <stdlib.h>#include <math.h>int main() { double a, b, c, discriminant, root1, root2, realPart, imagPart; printf("Enter coefficients a, b and c: "); scanf("%lf %lf %lf", &a, &b, &c); discriminant = b * b - 4 * a * c; if (discriminant > 0) { root1 = (-b + pow(discriminant,0.5)) / (2 * a); root2 = (-b - pow(discriminant,0.5)) / (2 * a); printf("root1 = %.2lf and root2 = %.2lf", root1, root2); } else if (discriminant == 0) { root1 = root2 = -b / (2 * a); printf("root1 = root2 = %.2lf;", root1); } else { printf("The equation has no real roots\n"); } return 0;}