#include <iostream>
#include <cmath>
#include <cstdio>

using namespace std;

int main()
{
    double a, b, c;
    cin >> a >> b >> c;

    if (b * b - (4 * c * a) < 0 || a == 0) cout << "Impossivel calcular" << endl;
    if (b * b - (4 * a * c) > 0 && a != 0) 
    printf("R1 = %.5lf\nR2 = %.5lf", (-b + sqrt(b * b - 4 * a * c )) / (2 * a), (-b - sqrt(b * b - 4 * a * c))/ (2 * a));

    return 0 ;
}

唯一一点就是注意a的问题：
a == 0和判别式 < 0时都为无解情况
有解时要注意有两个情况！！！