${x}^{2}+4x+3x+12=0$
Factor out $x$ from the expression$x \times \left( x+4 \right)+3x+12=0$
Factor out $3$ from the expression$x \times \left( x+4 \right)+3\left( x+4 \right)=0$
Factor out $x+4$ from the expression$\left( x+4 \right) \times \left( x+3 \right)=0$
When the product of factors equals $0$, at least one factor is $0$$\begin{array} { l }x+4=0,\\x+3=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=-4,\\x+3=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=-4,\\x=-3\end{array}$
The equation has $2$ solutions$\begin{array} { l }x_1=-4,& x_2=-3\end{array}$