Запишем область допустимых значений: \(2-x \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-x \ge -2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \le 2.\)
\({x^2}-2x + \sqrt {2-x} = \sqrt {2-x} + 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-2x = 3,\\x \le 2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-2x-3 = 0,\\x \le 2.\end{array} \right.\)
\({x^2}-2x-3 = 0;\,\,\,\,\,\,D = {\left( {-2} \right)^2}-4 \cdot 1 \cdot \left( {-3} \right) = 16;\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{2-4}}{2} = -1,\\x = \dfrac{{2 + 4}}{2} = 3.\end{array} \right.\)
\(\left\{ \begin{array}{l}{x^2}-2x-3 = 0,\\x \le 2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}x = -1,\\x = 3,\end{array} \right.\\x \le 2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -1.\)
Ответ: \(-1.\)