The one move is finding two numbers. For $x^{2} + bx + c$, look for two numbers that multiply to $c$ and add to $b$. In $x^{2} + 7x + 12$, the numbers $3$ and $4$ multiply to $12$ and add to $7$, so the factored form is $(x + 3)(x + 4) = 0$.

Then use the zero rule: if two things multiply to zero, one of them must be zero. Set each parenthesis equal to zero and solve. Here $x = -3$ and $x = -4$. Notice how the signs flip.