求一元二次方程的根

程序填空，不要改变与输入输出有关的语句。
输入一个正整数repeat (0<repeat<10)，做repeat次下列运算：
输入参数a,b,c，求一元二次方程a*x*x＋b*x＋c＝0的根，结果保留2位小数。
输出使用以下语句：
printf("参数都为零，方程无意义!\n");
printf("a和b为0，c不为0，方程不成立\n");
printf("x = %0.2f\n", -c/b);
printf("x1 = %0.2f\n", (-b+sqrt(d))/(2*a));
printf("x2 = %0.2f\n", (-b-sqrt(d))/(2*a));
printf("x1 = %0.2f+%0.2fi\n", -b/(2*a), sqrt(-d)/(2*a));
printf("x2 = %0.2f-%0.2fi\n", -b/(2*a), sqrt(-d)/(2*a));
输入输出示例：括号内为说明
输入：
5               (repeat=5)
0 0 0           (a=0,b=0,c=0)
0 0 1           (a=0,b=0,c=1)
0 2 4           (a=0,b=2,c=4)
2.1 8.9 3.5     (a=2.1,b=8.9,c=3.5)
1 2 3           (a=1,b=2,c=3)
输出：
参数都为零，方程无意义!
a和b为0，c不为0，方程不成立
x = -2.00
x1 = -0.44
x2 = -3.80
x1 = -1.00+1.41i
x2 = -1.00-1.41i

#include <stdio.h>
#include <math.h>
int main(void)
{
int repeat, ri;
double a, b, c, d;

scanf("%d", &repeat);
for(ri = 1; ri <= repeat; ri++){
scanf("%lf%lf%lf", &a, &b, &c);
d = pow(b,2)-4*a*c;
if(a==0 && b==0 && c==0)
printf("参数都为零，方程无意义!\n");
else if(a==0 && b==0 && c!=0)
printf("a和b为0，c不为0，方程不成立\n");
else if(a == 0)
printf("x = %0.2f\n", -c/b);
else if(d >= 0){
printf("x1 = %0.2f\n", (-b+sqrt(d))/(2*a));
printf("x2 = %0.2f\n", (-b-sqrt(d))/(2*a));
}
else {
printf("x1 = %0.2f+%0.2fi\n", -b/(2*a), sqrt(-d)/(2*a));
printf("x2 = %0.2f-%0.2fi\n", -b/(2*a), sqrt(-d)/(2*a));
}
}
}