đặt t = \(\sqrt{x^2+5x+10}\) t>0
\(t^2\)=\(x^2+5x+10\)
\(t^2-10\)=\(x^2+5x\)
thay vào pt ta đc
\(t^2\) -8+2t=0
\(\left\{{}\begin{matrix}t=2\left(tm\right)\\t=-4\left(l\right)\end{matrix}\right.\)
\(t^2\)=\(x^2+5x+10\)
\(x^2+5x+10\)=4
\(\left\{{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)