Question

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

Answer 1

We need two numbers `a` and `b` such that
`axxb=-8`
and
`a+b=-7`

The obvious (I hope) pair are `(a,b) = (-8,+1)`

and our factors are
`x^2-7x-8 = (x-8)(x+1)`