\(x^2+7x+12=0\)
\(\Leftrightarrow x^2+4x+3x+12=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)
Vậy .............
\(x^2+7x+12=0\)
\(x^2+3x+4x+12=0\)
\(x\left(x+3\right)+4\left(x+3\right)=0\)
\(\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)