Basic Quadratics

$x^2 + 3x + 12 = 10 \quad\Longrightarrow\quad x^2 + 3x + 2 = 0$

Factoring $\text{LHS}$ get us $(\color{#D61F06}{x+2})(\color{#3D99F6}{x+1}) = 0\quad\Longrightarrow\quad \color{#D61F06}{x+2} = 0 ~\text{ or }~ \color{#3D99F6}{x+1} = 0\quad\Longrightarrow\quad \color{#D61F06}{x} = \color{#D61F06}{-2} ~\text{ or }~ \color{#3D99F6}{x}=\color{#3D99F6}{-1}$ . Since $x$ cannot be equal to $-2$ , it leaves us $\boxed{x=-1}$

Subtracting 10 from both sides results in the equation $x^2 + 3x + 2 = 0$ which factors into $(x + 1)(x + 2) = 0$ , which has solutions $x = -1$ and $x = -2$ . However, we are told $x = -2$ is not allowed as a solution, so our only solution is $x = -1$ .