Решите неравенство: x^2+2x-8>0.

\[x^{2} + 2x - 8 > 0\]

\[x^{2} + 2x - 8 = 0\]

\[x_{1} + x_{2} = - 2;\ \ \ x_{1} \cdot x_{2} = - 8\]

\[x_{1} = - 4;\ \ \ \ \ x_{2} = 2\]

\[(x + 4)(x - 2) > 0\]

\[Ответ:( - \infty; - 4) \cup (2; + \infty).\]