ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow3x+2+2\sqrt{2x^2+x-3}=x+6\)
\(\Leftrightarrow\sqrt{2x^2+x-3}=2-x\) (\(x\le2\))
\(\Leftrightarrow2x^2+x-3=\left(2-x\right)^2\)
\(\Leftrightarrow x^2+5x-7=0\Rightarrow\left[{}\begin{matrix}x=\frac{-5-\sqrt{53}}{2}\left(l\right)\\x=\frac{-5+\sqrt{53}}{2}\end{matrix}\right.\)