# How do you solve x^2-7x-8<0?

We need two numbers $a$ and $b$ such that
$a \times b = - 8$
$a + b = - 7$
The obvious (I hope) pair are $\left(a , b\right) = \left(- 8 , + 1\right)$
${x}^{2} - 7 x - 8 = \left(x - 8\right) \left(x + 1\right)$