\(\sqrt {{x^3}-4{x^2}-10x + 29} = 3-x.\)
Уравнение вида \(\sqrt {f\left( x \right)} = g\left( x \right)\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{g\left( x \right) \ge 0,\;\;\;\;\;\,\;\;}\\{f\left( x \right) = {g^2}\left( x \right).}\end{array}} \right.\)
Тогда исходное уравнение примет вид:
\(\sqrt {{x^3}-4{x^2}-10x + 29} = 3-x\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ \begin{array}{l}3-x \ge 0,\\{x^3}-4{x^2}-10x + 29 = 9-6x + {x^2}\end{array} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left\{ \begin{array}{l}x \le 3,\\{x^3}-5{x^2}-4x + 20 = 0\end{array} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ \begin{array}{l}x \le 3,\\{x^2}\left( {x-5} \right)-4\left( {x-5} \right) = 0\end{array} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ \begin{array}{l}x \le 3,\\\left( {x-5} \right)\left( {{x^2}-4} \right) = 0\end{array} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left\{ \begin{array}{l}x \le 3,\\\left[ {\begin{array}{*{20}{c}}{x-5 = 0,\;}\\{{x^2}-4 = 0}\end{array}} \right.\end{array} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ \begin{array}{l}x \le 3,\\\left[ {\begin{array}{*{20}{c}}{x = 5,\;\;}\\{\begin{array}{*{20}{c}}{x = -2,}\\{x = 2\;\;\;}\end{array}}\end{array}} \right.\end{array} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -2,}\\{x = 2.\;\;}\end{array}} \right.\)
ОТВЕТ: -2; 2.