我自己编了一个求解啊a*x*x+b*x+c=0程序

#include "iostream.h"
#include <cmath>
int main()
{
    double x,y,a,b,c,w,u,dlt;
int k,l, m;
cout<<"求解形如a*x*x+b*x+c=0的方程程序："<<endl;   
   
{ cout<<"请输入a"<<"  ";
    cin>>a;
    cout<<"是否对a开方,若是请输入k==1,否则输入其他值"<<" k=";  
    cin>>k;
   if(k==1)
       a=sqrt(a);
   else
      
       a=a;
    }
    {cout<<"请输入b"<<"  ";
    cin>>b;
    cout<<"是否对b开方若是,请输入l==1,否则输入其他值"<<" l=";  
    cin>>l;
   if(l==1)
       b=sqrt(b);
   else
     
       b=b;
    }
    { cout<<"请输入c"<<"  ";
    cin>>c;
    cout<<"是否对c开方,若是请输入m==1,否则输入其他值"<<" m=";  
    cin>>m;
   if(m==1)
       c=sqrt(c);
   else
       c=c;


    }
  if(a==0)
  { cout<<"方程是一元一次方程"<<endl;
    x=-c/b;
     cout<<"x="<<x<<endl;
    return 0;
  }

   else
   dlt=b*b-4*a*c;
   if(dlt==0)
    {x=(-b+sqrt(dlt))/2/a;
     cout<<"方程有等实根"<<endl;
    cout<<"x=y="<<x<<endl;

    }
    else
    if(dlt>0)
       {cout<<"方程有两个不相等的实根"<<endl;
   
        x=(-b+sqrt(dlt))/2/a;
        y=(-b-sqrt(dlt))/2/a;
      
        cout<<"x="<<x<<endl;
        cout<<"y="<<y<<endl;
    }

         else
         {w=(-b)/2/a;
        u=sqrt(fabs(dlt))/2/a;   

        cout<<"方程有虚根"<<endl;

        cout<<"x="<<w<<"+"<<u<<"i"<<endl;
        cout<<"y="<<w<<"-"<<u<<"i"<<endl;
         }
}