\(x^2-2x=5\Leftrightarrow x^2-2x+1=6\Leftrightarrow\left(x-1\right)^2=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{6}\\x-1=-\sqrt{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{6}+1\\x=-\sqrt{6}+1\end{matrix}\right.\)
Vậy \(x\in\left\{\sqrt{6}+1;-\sqrt{6}+1\right\}\)