c语言求方程ax²+bx+c=0的根，用3个函数分别求当：b²-4ac大于0，等于0，小于0时的根并输出结果。从主函数输入a,b,c的值

以下是C语言求解一元二次方程的程序，其中包括三个函数分别求当b²-4ac大于0，等于0，小于0时的根并输出结果。从主函数输入a,b,c的值：

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

void roots_greater_than_zero(float a, float b, float c);
void roots_equal_to_zero(float a, float b, float c);
void roots_less_than_zero(float a, float b, float c);

int main() {
    float a, b, c;
    printf("Please enter the value of a, b, c:\n");
    scanf("%f%f%f", &a, &b, &c);
    float discriminant = b * b - 4 * a * c;
    if (discriminant > 0) {
        roots_greater_than_zero(a, b, discriminant);
    } else if (discriminant == 0) {
        roots_equal_to_zero(a, b);
    } else {
        roots_less_than_zero(a, b, discriminant);
    }
    return 0;
}

void roots_greater_than_zero(float a, float b, float discriminant) {
    float root1 = (-b + sqrt(discriminant)) / (2 * a);
    float root2 = (-b - sqrt(discriminant)) / (2 * a);    printf("The roots are real and different.\n");
    printf("root1 = %.2f, root2 = %.2f\n", root1, root2);
}

void roots_equal_to_zero(float a, float b) {
    float root = -b / (2 * a);
    printf("The roots are real and equal.\n");
    printf("root1 = root2 = %.2f\n", root);
}

void roots_less_than_zero(float a, float b, float discriminant) {
    float realPart = -b / (2 * a);
    float imaginaryPart = sqrt(-discriminant) / (2 * a);
    printf("The roots are complex and different.\n");
    printf("root1 = %.2f + %.2fi, root2 = %.2f - %.2fi\n", realPart, imaginaryPart, realPart, imaginaryPart);
}