1.4-20:求一元二次方程的根

#include 
#include 
#include  
using namespace std;
int main()
{
    double a,b,c,x;
    cin>>a>>b>>c;
    if(b*b==4*a*c)
        printf("x1=x2=%.5f",(-b+sqrt(b*b-4*a*c))/(2*a));
    else
    {
        if(b*b>4*a*c)
            printf("x1=%.5f;x2=%.5f",(-b+sqrt(b*b-4*a*c))/(2*a), (-b-sqrt(b*b-4*a*c))/(2*a));
        else
        {
            x=(-b/(2*a));
            if(x==-0.00000)
                x=0;
            printf("x1=%.5f+%.5fi;x2=%.5f-%.5fi",x,(sqrt(4*a*c-b*b)/(2*a)),x,(sqrt(4*a*c-b*b)/(2*a)));
        }
    }
    return 0;
}