\[x^{2} + 3x + 2 > 0\]
\[x_{1} + x_{2} = - 3;x_{1} \cdot x_{2} = 2\]
\[x_{1} = - 2;\ \ x_{2} = - 1;\]
\[(x + 2)(x + 1) > 0\]
\[x < - 2;\ \ x > - 1.\]
\[Ответ:x < - 2;x > - 1.\]