\(x\left(x+1\right)=12\Leftrightarrow x^2+x=12\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow x^2-3x+4x-12=0\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\Leftrightarrow\left\{{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\x=3\end{matrix}\right.\)
vậy \(x=-4;x=3\)