#include<iostream>
#include<set>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<map>
#include<numeric>
#include<queue>
#include<cmath>
using namespace std;
typedef long long ll;
const ll INF=1LL<<60;
typedef pair<ll,ll> P;
typedef pair<int,P> PP;
const ll MOD=1e9+7;

ll gcd(ll x,ll y){
    return y==0?x:gcd(y,x%y);
}

double f(ll n){
    const int N=100;
    double lb=0,ub=n+1;
    for(int t=0;t<N;t++){
        double mid=(lb+ub)/2;

        if(mid*mid<=n){
            lb=mid;
        }else{
            ub=mid;
        }
    }

    return lb;
}

void print(double x){
    printf("%.23f\n",x);
}

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

    if(a==0){

        if(b==0){
            if(c==0){
                //a=b=c=0
                //解xは∞個
                cout<<-1<<endl;
            }else{
                //a=b=0, c!=0
                //解なし
                cout<<0<<endl;
            }
        }else{
            //b!=0
            // bx+c=0
            cout<<1<<endl;

            // ll g=gcd(b,c);
            // b/=g;
            // c/=g;

            {
                ll g=1;
                
                for(ll t=1;t*t<=min({a,b,c});t++ ){

                    if(b%g==0 && c%g==0){
                        g=t;
                    }

                }

                b/=g;
                c/=g;

            }



            double x=-1.0*c/b;
            //printf("%.12f\n",x);
            print(x);


        }
        
        return 0;
    }

    //ll g=gcd(c,gcd(b,a));
    {
        ll g=1;
        
        for(ll t=1;t*t<=min({a,b,c});t++ ){

            if(a%g==0 && b%g==0 && c%g==0){
                g=t;
            }

        }


        a/=g;
        b/=g;
        c/=g;

    }

    if(a<0){
        a*=(-1);
        b*=(-1);
        c*=(-1);
    }

    //a!=0

    ll D=b*b-4*a*c;

    if(D==0){
        cout<<1<<endl;
        double x=1.0*(-b)/(2*a);
        //printf("%.12f\n",x);
        print(x);


    }else if(D>0){
        double x1=(1.0*(-b)-sqrt(D))/(2*a);
        double x2=(1.0*(-b)+sqrt(D))/(2*a);

        //double x1=(1.0*(-b)-f(D))/(2*a);
        //double x2=(1.0*(-b)+f(D))/(2*a);


        x1=min(x1,x2);
        x2=max(x1,x2);
        cout<<2<<endl;
        //printf("%.12f\n",x1);
        //printf("%.12f\n",x2);
        print(min(x1,x2));
        print(max(x1,x2));

    }else{
        //D<0
        //虚数解
        cout<<0<<endl;
    }
}