Запишем область допустимых значений: \(5-x \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-x \ge -5\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \le 5.\)
\({x^2}-2x + \sqrt {5-x} = \sqrt {5-x} + 24\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-2x = 24,\\x \le 5\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-2x-24 = 0,\\x \le 5.\end{array} \right.\)
\({x^2}-2x-24 = 0;\,\,\,\,\,\,D = {\left( {-2} \right)^2}-4 \cdot 1 \cdot \left( {-24} \right) = 100;\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{2-10}}{2} = -4,\\x = \dfrac{{2 + 10}}{2} = 6.\end{array} \right.\)
\(\left\{ \begin{array}{l}{x^2}-2x-24 = 0,\\x \le 5\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}x = -4,\\x = 6,\end{array} \right.\\x \le 5\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -4.\)
Ответ: \(-4.\)