#include<stdio.h>
#include<math.h>
int main()
{
    double a,b,c,d,e,x1,x2;
    while(scanf("%lf%lf%lf",&a,&b,&c)!=EOF)
    {   
    d=b*b-4*a*c;
    if(d>0)
    {
    e=sqrt(d);
    x1=(-b+e)/(2*a);
    x2=(-b-e)/(2*a);
    printf("x1=%.3f,x2=%.3f\n",x1,x2);
    }
    else if(d=0)
    {
    x1=(-b)/(2*a);
    printf("x1=x2=%.3f\n",x1);
    }
    else
    {
    e=sqrt(-d)
    x1=(-b+i*e)/(2*a);
    x2=(-b-i*e)/(2*a);
    printf("x1=%.3f,x2=%.3f\n",x1,x2);
    }
    }
return 0;
}

去掉i那不是不对了吗，求解释

试了，但结果是
编译错误信息：

--------------------------------------------------------------------------------

Main.cpp: In function 'int main()':
Main.cpp:16: warning: suggest parentheses around assignment used as truth value
Main.cpp:24: error: expected `:' before ';' token
Main.cpp:24: error: expected primary-expression before ';' token

这样啊

#include<stdio.h>
#include<math.h>
int main()
{
    double a,b,c,d,e,x1,x2,m,n;
    while(scanf("%lf%lf%lf",&a,&b,&c)!=EOF)
    {
        d=b*b-4*a*c;
        if(d>0)
        {
        e=sqrt(d);
        x1=(-b+e)/(2*a);
        x2=(-b-e)/(2*a);
        printf("x1=%.3f,x2=%.3f\n",x1,x2);
        }
        else if(d<0)
        {
            e=sqrt(-d);
            m=-b/(2*a);
            n=e/(2*a);
            printf("x1=%.3f+%.3fi,x2=%.3f-%.3fi\n",m,n,m,n);
        }
        else
        {
            x1=(-b)/(2*a);
            printf("x1=x2=%.3f\n",x1);
        }
        
    }
        return 0;
}
那先在呢？我还是通不过啊

额，是x1x2之间的空格没弄
谢了啊