Đặt \(\sqrt{2x^2+3x+2}=t>0\)
\(\Rightarrow4x^2+6x+21=2t^2+17\)
Phương trình trở thành:
\(t+\sqrt{2t^2+17}=11\Leftrightarrow\sqrt{2t^2+17}=11-t\)
\(\Leftrightarrow\left\{{}\begin{matrix}11-t\ge0\\2t^2+17=\left(11-t\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\le11\\t^2+22t-104=0\end{matrix}\right.\)
\(\Rightarrow t=4\Leftrightarrow2x^2+3x+2=16\)
\(\Leftrightarrow2x^2+3x-14=0\)
\(\Leftrightarrow...\)