Решите неравенство x^2-4x-5>0.

\[x^{2} - 4x - 5 > 0\]

\[x^{2} - 4x - 5 = 0\]

\[x_{1} + x_{2} = 4;\ \ x_{1} \cdot x_{2} = - 5\]

\[x_{1} = 5;\ \ x_{2} = - 1.\]

\[(x + 1)(x - 5) > 0\]

\[x \in ( - \infty; - 1) \cup (5; + \infty).\]