求解！C++编程题目

求方程ax² +bx+c=0的根，分别求当b²-4ac大于0、等于0和小于0时的根，并输出结果。a,b,c的值从键盘输入。

#include<iostream.h>
#include<math.h>
void main()
{
double a,b,c,temp,x1,x2;
cin>>a>>b>>c;
temp=b*b-4*a*c;
if(temp<0)
{
temp=sqrt(-1*temp);
x1=(-1*b)/(2*a);
x2=temp/(2*a);
cout<<"方程有两复根"<<endl;
cout<<"x1="<<x1<<"+"<<x2<<"i"<<endl;
cout<<"x2="<<x1<<"-"<<x2<<"i"<<endl;
}
else
{
x1=(-1*b)/(2*a);
if(temp==0)
{
cout<<"方程有两相同实根"<<endl;
cout<<"x1=x2="<<x1<<endl;
}
else
{
x2=temp/(2*a);
cout<<"方程有两不同实根"<<endl;
cout<<"x1="<<x1+x2<<endl;
cout<<"x2="<<x1-x2<<endl;
}
}
}

#include "iostream" #include "math.h" using namespace std; int main(){ float a,b,c; float x1=1,x2=1; cin>>a>>b>>c; float delt = b*b-4*a*c; if (delt<0) { cout<<"x1="<<-b/(2*a)<<"+"<<sqrt(-delt)/(2*a)<<"i"; cout<<"x2="<<-b/(2*a)<<"-"<<sqrt(-delt)/(2*a)<<"i"; } else if(delt>0) { x1 = (-b+sqrt(delt))/(2*a); x2 = (-b-sqrt(delt))/(2*a); cout<<"x1="<<x1<<" ,x2="<<x2; } else { cout<<"x1=x2="<<-b/(2*a); } }