\(\sqrt{x+6}=x\)
\(\Rightarrow x+6=x^2\)
\(\Leftrightarrow x+6-x^2=x^2-x^2\)
\(\Leftrightarrow-x^2+x+6=0\)
\(\Leftrightarrow\left(-x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-2=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
Vậy \(x=3\)