Đặt \(t=x^2+x\) ta có pt trở thành:
\(t^2+4t=12\Leftrightarrow t^2+4t-12=0\)
\(\Leftrightarrow t^2-2t+6t-12=0\)\(\Leftrightarrow t\left(t-2\right)+6\left(t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+6\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=2\\x^2+x=-6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)\left(x-1\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
\(\Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)+4=8\)
\(\Leftrightarrow\left(x^2+x+2\right)^2=8\)
\(\left[{}\begin{matrix}x^2+x+2=8\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\x^2+x+2=-8\left(loai\right)\end{matrix}\right.\)
