C语言解法 自定义函数求一元二次方程

解题思路:正常思路    

注意事项:符号 和 括号

参考代码:

#include<stdio.h>

#include<math.h>

void yuan(int a, int b, int c)

{

double x1, x2, u, k;

double d = ((double)b * b) - ((double)4 * a * c);

if (d > 0) {

x1 = (-b + fabs(sqrt(d))) / ((double)2 * a);

x2 = (-b - fabs(sqrt(d))) / ((double)2 * a);

printf("x1=%.3lf x2=%.3lf", x1, x2);

}

else if (d == 0) {

x1 = x2 = (-2 * a) / b;

printf("x1=%.3lf x2=%.3lf", x1, x2);

}

else if (d < 0) {

d = sqrt(-d);

x1 = x2 = -b / ((double)2 * a);

u = k= d / ((double)2 * a);

printf("x1=%.3lf+%.3lfi x2=%.3lf-%.3lfi", x1, u, x2, k);

}

}

int main() {

int a, b, c;

scanf("%d %d %d", &a, &b, &c);

yuan(a, b, c);

return 0;

}