#include <bits/stdc++.h> 
using namespace std;
int T, m, a, b, c, d, k, t, n, e;
void Main() {
	cin >> a >> b >> c;
	if(a < 0)
		a = -a, b = -b, c = -c;
	d = b * b - 4 * a * c, k = 1;
	if(d < 0) {
		cout << "NO\n";
		return;
	}
	for(int i = 2; i * i <= d; i++)
		while(d % (i * i) == 0)
			k *= i, d /= i * i;
	if(d == 0 || d == 1){
		t = abs(__gcd(2 * a , -b + k * d));
		cout << (-b + k * d) / t;
		if(2 * a / t != 1)
			cout << '/' << 2 * a / t;
		cout << '\n';
		return;
	}
	t = abs(__gcd(__gcd(-b, 2 * a), k));
	b /= -t;
	a = 2 * a / t;
	k /= t;
	n = abs(__gcd(a, b));
	a /= n;
	b /= n;
	if(b){
		if(b % a == 0) cout << b / a;
		else cout << b << '/' << a;
		cout << '+';
	}
	if(abs(k) == 1)
	cout << "sqrt(" << d << ")"; 
	else cout << abs(k) << "*sqrt(" << d << ")"; 
	if(a == 1) cout << '\n';
	else cout << '/' << a << '\n';
	return;
}
int main() {
	//freopen("P9750_9.in", "r", stdin);
	//freopen("157.txt", "w", stdout);
	for(cin >> T >> m; T; --T) Main();
	return 0;
}

第9, 10个数据点过不了

原题