题解 | #计算一元二次方程#

#include<iomanip>
#include<cmath>
using namespace std;
int main()
{
    float a,b,c;
    float d,e;
    while(cin>>a>>b>>c)
    {
        e=-b/(2*a);
        d=(b*b)-(4*a*c);
        if(a==0)
        cout<<"Not quadratic equation"<<endl;
      else if(a!=0)
      {
        if(d==0)
        {
            cout << fixed << setprecision(2);
            cout<<"x1=x2="<<e<<endl;
        }
        else if(d>0)
        {
            cout << fixed << setprecision(2);
            cout<<"x1="<<e-sqrt(d)/(2*a)<<";"<<"x2="<<e+sqrt(d)/(2*a)<<endl;
        }
        else
        {
            cout << fixed << setprecision(2);
            cout<<"x1="<<e<<"-"<<sqrt(fabs(d))/(2*a)<<"i"<<";"<<"x2="<<e<<"+"<<sqrt(fabs(d))/(2*a)<<"i"<<endl;
        }
      }
    }
    return 0;
}

sqrt里面必须是正数derta小于零时：-1开方=i，里面按正数算