f（X）＝aX²＋bX－c的求根完整代码

f（X）＝aX²＋bX－c的求根完整代码的

尽量少点代码

#include<bits/stdc++.h>
using namespace std;
int a,b,c;
int main(){
    scanf("%d%d%d",&a,&b,&c);
    if(b*b-4*a*c>0){//有两个解 
        cout<<(-b+sqrt(b*b-4*a*c))/(2*a)<<" "<<(-b-sqrt(b*b-4*a*c))/(2*a);
    }else if(b*b-4*a*c==0){//有唯一解 
        cout<<b/(2*a); 
    }else{
        cout<<"无解"; 
    }
    return 0;
}

望采纳

#include<bits/stdc++.h>
using namespace std;
int main()
{
    double a,b,c,x1,x2,t;
    cin>>a>>b>>c;
    t=b*b-4*a*c;
    if(t>0){          //如果有解
        x1=(-b-sqrt(t))/(2*a);
        x2=(-b+sqrt(t))/(2*a);
        cout<<"x1="<<x1<<","<<"x2="<<x2;
    }
    else if(t==0)
        cout<<"x1=x2="<<-b/(2*a);  //x1+x2=-b/a,并且两根相同所以两根相等x1=x2=-b/(2*a)
    else
        cout<<"无解";
    return 0;
}