#include<iostream>
#include<cmath>
using namespace std;

/*
有解：b*b-4*a*c>=0
    解：(-b+sqrt(b*b-4*a*c))/2*a
       (-b-sqrt(b*b-4*a*c))/2*a
无解：b*b-4*a*c<0
*/

int t;
double a,b,c;
double x1,x2;

int main()
{
    cin>>t;
    while(t--)
    {
        scanf("%lf %lf %lf",&a,&b,&c);
        double d=b*b-4*a*c;
        if(d<0)
        printf("-1\n");
        else
        {
            x1=(-b-sqrt(d))/2.0*a;
            x2=(-b+sqrt(d))/2.0*a;
            if(x1!=x2)
            printf("x1=%.2f,x2=%.2f\n",x1,x2);
            else
            printf("x=%.2f\n",x1);
        }
    }
    return 0;
}