ĐKXĐ: \(x^2+5x+2\ge0\)
\(x^2+5x+4-3\sqrt{x^2+5x+2}-6=0\)
Đặt \(\sqrt{x^2+5x+2}=a\ge0\)
\(\Rightarrow a^2-3a-4=0\Rightarrow\left[{}\begin{matrix}a=-1< 0\left(l\right)\\a=4\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+5x+2}=4\Leftrightarrow x^2+5x+2=16\)
\(\Leftrightarrow x^2+5x-14=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)