Запишем область допустимых значений: \(4-x \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-x \ge -4\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \le 4.\)
\({x^2}-2x + \sqrt {4-x} = \sqrt {4-x} + 15\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-2x = 15,\\x \le 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-2x-15 = 0,\\x \le 4.\end{array} \right.\)
\({x^2}-2x-15 = 0;\,\,\,\,\,\,D = {\left( {-2} \right)^2}-4 \cdot 1 \cdot \left( {-15} \right) = 64;\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{2-8}}{2} = -3,\\x = \dfrac{{2 + 8}}{2} = 5.\end{array} \right.\)
\(\left\{ \begin{array}{l}{x^2}-2x-15 = 0,\\x \le 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}x = -3,\\x = 5,\end{array} \right.\\x \le 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -3.\)
Ответ: \(-3.\)