Запишем область допустимых значений: \(6-x \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-x \ge -6\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \le 6.\)
\({x^2}-6x + \sqrt {6-x} = \sqrt {6-x} + 7\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-6x = 7,\\x \le 6\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-6x-7 = 0,\\x \le 6.\end{array} \right.\)
\({x^2}-6x-7 = 0;\,\,\,\,\,\,D = {\left( {-6} \right)^2}-4 \cdot 1 \cdot \left( {-7} \right) = 64;\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{6-8}}{2} = -1,\\x = \dfrac{{6 + 8}}{2} = 7.\end{array} \right.\)
\(\left\{ \begin{array}{l}{x^2}-6x-7 = 0,\\x \le 6\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}x = -1,\\x = 7,\end{array} \right.\\x \le 6\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -1.\)
Ответ: \(-1.\)