\[x^{2} + 6x + 5 > 0\]
\[D = 6^{2} - 4 \cdot 1 \cdot 5 =\]
\[= 36 - 20 = 16\]
\[x_{1} = \frac{- 6 + \sqrt{16}}{2 \cdot 1} = \frac{- 6 + 4}{2} =\]
\[= - \frac{2}{2} = - 1;\ \ \ \ \ \]
\[x_{2} = \frac{- 6 - \sqrt{16}}{2 \cdot 1} = \frac{- 6 - 4}{2} =\]
\[= - \frac{10}{2} = - 5\]
\[Ответ:( - \infty; - 5) \cup ( - 1; + \infty).\]