\[\boxed{\text{504}\text{\ (504)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ (x + 2)^{2} + 9 \cdot (x + 2) + 20 =\]
\[= 0\]
\[Пусть\ t = x + 2,\ \]
\[тогда\ \ t^{2} + 9t + 20 = 0.\]
\[По\ теореме\ Виета:\ \ t_{1} = - 5,\ \ \]
\[t_{2} = 4.\]
\[1)\ x + 2 = - 5 \Longrightarrow x_{1} = - 7.\]
\[2)\ x + 2 = - 4 \Longrightarrow x_{2} = - 6.\]
\[\textbf{б)}\ (x - 5)^{2} + 2 \cdot (x - 5) - 63 =\]
\[= 0\]
\[Пусть\ t = x - 5,\]
\[тогда\ \text{\ \ }t^{2} + 2t - 63 = 0.\]
\[По\ теореме\ Виета:t_{1} = - 9,\ \ \]
\[t_{2} = 7.\]
\[1)\ x - 5 = - 9 \Longrightarrow x_{1} = - 4.\]
\[2)\ x - 5 = 7 \Longrightarrow x_{2} = 12.\]
\[Ответ:а) - 7;\ - 6;\ \ б) - 4;12.\]