Запишем область допустимых значений: \(3-x \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-x \ge -3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \le 3.\)
\({x^2}-2x + \sqrt {3-x} = \sqrt {3-x} + 8\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-2x = 8,\\x \le 3\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-2x-8 = 0,\\x \le 3.\end{array} \right.\)
\({x^2}-2x-8 = 0;\,\,\,\,\,\,D = {\left( {-2} \right)^2}-4 \cdot 1 \cdot \left( {-8} \right) = 36;\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{2-6}}{2} = -2,\\x = \dfrac{{2 + 6}}{2} = 4.\end{array} \right.\)
\(\left\{ \begin{array}{l}{x^2}-2x-8 = 0,\\x \le 3\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}x = -2,\\x = 4,\end{array} \right.\\x \le 3\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -2.\)
Ответ: \(-2.\)