一元二次方程ax2+bx+c=0

判別式D=b2−4ac
一元二次方程的實根：x=−b±D√2a
一元二次方程的虛根：x=−b2a±D√2ai

用switch語句實現：

#include "stdio.h"
#include "math.h"

int main(void)
{
    float a = 0, b = 0, c = 0;
    float discriminant = 0, x1 = 0, x2 = 0;
    float p = 0, q = 0;

    printf("Please input three coefficient:");
    scanf_s("%f %f %f"
 
, &a, &b, &c);

    discriminant = pow(b, 2) - 4 * a * c;

    switch (discriminant >= 0)
    {
    case 1:
        switch (discriminant > 0)
        {
        case 1:
            x1 = (-b + sqrt(discriminant)) / (2 * a);
            x2 = (-b - sqrt(discriminant)) / (2 * a);
            printf
 
("兩個不相等的實根為：x1 = %f, x2 = %f\n", x1, x2);
            break;
        case 0:
            x1 = -b / (2 * a);
            printf("兩個相等的實根為:x1 = x2 = %f\n", x1);
            break;
        }
        break;
    case 0:
        p = -b / (2 * a);
        q = sqrt(-discriminant) / (2 * a);
        printf("兩個不相等的虛根為：x1 = %f + %fi, x2 = %f - %fi\n"
 
, p, q, p, q);
        break;
    }

    return 0;
}