\(\sqrt{5x^2+10x+1}=-\frac{1}{5}\left(5x^2+10x+1\right)+\frac{36}{5}\)
Đặt \(\sqrt{5x^2+10x+1}=t\left(t\ge0\right)\)
\(pt\Leftrightarrow t=-\frac{1}{5}t^2+\frac{36}{5}\)
\(\Leftrightarrow-t^2+36-5t=0\)
\(\Rightarrow t=4\left(tm\right)\)
\(\Leftrightarrow5x^2+10x+1=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)