关于二元一次方程

#include <math.h>
#include <stdio.h>
#include <iostream.h>
void main()
{
    double a,b,c,d,x1,x2;  
    cin>>a>>b>>c;
    d=b*b-4*a*c;
    {
        if(d>=0)
    {
        x1=(-b+sqrt(d))/(2*a);
        x2=(-b-sqrt(d))/(2*a);
        printf("%.2f\n%.2f\n",x1,x2);   //前面定义的双精度型，这里为何用整型
//      cout<<"x1="<<x1<<"\t"<<"x2="<<x2<<endl;
    }
        else
        printf("无解\n");
//      cout<<"无解"<<endl;
    }

}