#include <bits/stdc++.h>
#include <iostream>
#include <limits>
#include <numeric>
#include <type_traits>
#include <bitset>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <list>

using namespace std;

#define rep(i,n,m) for(ll (i)=(n);(i)<(m);(i)++)
#define rrep(i,n,m) for(ll (i)=(n);(i)>(m);(i)--)
using ll = long long;
using ld = long double;
const ll mod = 1000000007;
const ll inf = 1000000000;
const ll INF = 1e15;


void pline(vector<ll> lis){
    rep(i,0,lis.size()){
        printf ("%lld",lis[i]);
        if (i != lis.size()-1) printf(" ");
        else printf("\n");
    }
}

void plineendl(vector<ll> lis){
    rep(i,0,lis.size()){
        cout << lis[i] << " ";
    }
    cout << endl;
}


ll llmin(ll x , ll y){
    if ( x < y ) return x;
    else return y;
}

vector<ld> solve(ll a, ll b, ll c){
    
    vector<ld> ret(0);

    if (a == 0 && b == 0 && c == 0){
        ret.push_back(INF);
    }else if (a == 0 && b == 0){
        //なにもしない
    }else if (a == 0){
        ret.push_back(-(ld)c/ (ld)b);
    }else{
        ll kai = b*b - 4*a*c;
        if (kai < 0){
            //なにもしない
        }else if (kai == 0){
            ret.push_back(-b/((ld)2*a));
        }else{
            ld sq = sqrt((double)kai);
            ret.push_back( (-b+sq)/(2*a) );
            ret.push_back( (-b-sq)/(2*a) );
        }
    }

    return ret;
}

int main(){

    ll a,b,c;
    scanf("%lld%lld%lld",&a,&b,&c);
    auto ans = solve(a,b,c);
    
    if (ans.size() == 0){
        cout << 0 << endl;
    }else if (ans[0] == INF){
        cout << -1 << endl;
    }else{
        sort(ans.begin(),ans.end());
        cout << ans.size() << endl;
        rep(i,0,ans.size()){
            cout << std::setprecision(13) << ans[i] << endl;
        }
    }

}

/*
int main(){

    ll N,R;
    scanf("%lld%lld",&N,R);

    vector<vector<ll>> xy;
    rep(i,0,N){
        ll x,y;
        scanf("%lld%lld",&x,&y);
        xy.push_back({x,y});
    }

    //線分はいいが、内外判定がきついな
    rep(i,0,N){
        rep
    }

    vector<vector<ld>> stime(N+1,vector<ld> (N+1,1e9));
    vector<vector<ld>> etime(N+1,vector<ld> (N+1,1e9));

    rep(i,0,N){
        rep(j,0,i){
            
            ll X = P[j][0]- P[i][0];
            ll Y = P[j][1]- P[i][1];
            ll VX = P[j][2]-P[i][2];
            ll VY = P[j][3]-P[i][3];
            
            // まず、直線の式を算出
            //時刻 tが直接二次方程式の解になる

            vector<ld> time = solve(vx*vx+vy*vy, 2*x*vx+2*y*vy, x*x+y*y-D*D);

            printf ("vertices = %lld %lld\n",i,j);
            pline(time);

        }
    }
    
}
*/

/*

*/