Запишем область допустимых значений: \(5-x \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-x \ge -5\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \le 5.\)
\({x^2}-3x + \sqrt {5-x} = \sqrt {5-x} + 18\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-3x = 18,\\x \le 5\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-3x-18 = 0,\\x \le 5.\end{array} \right.\)
\({x^2}-3x-18 = 0;\,\,\,\,\,\,D = {\left( {-3} \right)^2}-4 \cdot 1 \cdot \left( {-18} \right) = 81;\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{3-9}}{2} = -3,\\x = \dfrac{{3 + 9}}{2} = 6.\end{array} \right.\)
\(\left\{ \begin{array}{l}{x^2}-3x-18 = 0,\\x \le 5\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}x = -3,\\x = 6,\end{array} \right.\\x \le 5\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -3.\)
Ответ: \(-3.\)