Solve the inequation: $\;$ $x^2 - 14x - 15 > 0$
Given inequation: $\;$ $x^2 - 14x - 15 > 0$
i.e. $\;$ $\left(x - 15\right) \left(x + 1\right) > 0$
i.e. $\;$ $x - 15 > 0 \; \cap \; x + 1 > 0$ $\;$ OR $\;$ $x - 15 < 0 \; \cap \; x + 1 < 0$
i.e. $\;$ $x > 15 \; \cap \; x > -1$ $\implies$ $x > 15$
OR $\;$ $x < 15 \; \cap \; x < -1$ $\implies$ $x < -1$
i.e. $\;$ $x \in \left(15, + \infty\right)$ $\;$ OR $\;$ $x \in \left(- \infty, -1\right)$
$\therefore \;$ The solution is $\;$ $x \in \left(- \infty, -1\right) \cup \left(15, + \infty\right)$