Запишем область допустимых значений: \(6-x \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-x \ge -6\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \le 6.\)
\({x^2}-2x + \sqrt {6-x} = \sqrt {6-x} + 35\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-2x = 35,\\x \le 6\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-2x-35 = 0,\\x \le 6.\end{array} \right.\)
\({x^2}-2x-35 = 0;\,\,\,\,\,\,D = {\left( {-2} \right)^2}-4 \cdot 1 \cdot \left( {-35} \right) = 144;\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{2-12}}{2} = -5,\\x = \dfrac{{2 + 12}}{2} = 7.\end{array} \right.\)
\(\left\{ \begin{array}{l}{x^2}-2x-35 = 0,\\x \le 6\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}x = -5,\\x = 7,\end{array} \right.\\x \le 6\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -5.\)
Ответ: \(-5.\)