\[\boxed{\text{346.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[x^{2} + 6x - 7 \leq 0\]
\[D = 3^{2} + 7 = 16\]
\[x_{1,2} = - 3 \pm 4\]
\[(x + 7)(x - 1) \leq 0 \Longrightarrow\]
\[\Longrightarrow x \in \lbrack - 7;1\rbrack;\]
\[x^{2} - 2x - 15 \leq 0\]
\[D = 1 + 15 = 16\]
\[x_{1,2} = 1 \pm 4\]
\[(x + 3)(x - 1) \leq 0 \Longrightarrow\]
\[\Longrightarrow x \in \lbrack - 3;1\rbrack.\]
\[Общее\ решение:\ x \in \lbrack - 3;1\rbrack.\]