This short code allows you to enter B and C in a trinomial and then outputs it in factored form, and gives you the solutions.

Note: this only works if a = 1
(ax^2 + Bx + C)
Example:
Enter B: 7
Enter C: 10
Factored Form: (x+5)(x+2)
Solutions: x=-5, x=-2

Code:
#include<iostream.h>
#include<vector>
using namespace std;
int check(vector<int> vec, int b, int c)
{
    for(vector<int>::iterator x = vec.begin(); x!=vec.end(); x++)
    {
          for(vector<int>::iterator y = vec.end() - 1; y!=vec.begin(); y--)
          {
               if((*x + *y) == b && (*x)*(*y)== c)
               {
                     cout<<"Factored form is: "<<"(x ";
                     if(*x>0)
                             cout<<"+ "<<*x<<")(x ";
                     else
                             cout<<*x<<")(x ";
                     if(*y>0)
                             cout<<"+ "<<*y<<")";
                     else
                             cout<<*y<<") ";
                     cout<<endl<<"Solutions: "<<"x = "<<-1*(*x)<<", x = "<<-1*(*y);
                     return 1;
               }
          }
    }
    cout<<"This problem can't be solved. No factors of C add up to B"<<endl;
    return 0;
}
int main()
{
    vector<int> vec;
    int b, c;
    cout<<"This can solve equations in the following format:"<<endl;
    cout<<"x^2 + Bx + C   "<<endl;
    cout<<"Enter B: ";
    cin>>b;
    cout<<"Enter C: ";
    cin>>c;
    for(int a = c * -1; a<c; a++)  
    {
         if(c%a == 0)
         {
               vec.push_back(a);
         }
         if(a==(-1))//to make it skip zero. 
               a++;
    }
    check(vec, b, c);
    cin.get();
    cin.get();
    return 0;
}