Between the Roots

To solve ${x}^{2}-4x-5 < 0$, factor first: $\left(x-5\right)\left(x+1\right) < 0$. Now the question is about signs. A product of two numbers is negative exactly when one factor is positive and the other is negative, so there are two sign cases.

Case one: $x-5 > 0$ and $x+1 < 0$, so x is above 5 and below -1 at once. No number does both, so this case is empty. Case two: $x-5 < 0$ and $x+1 > 0$, so x is below 5 and above -1. That holds for every number between the roots, written as one chain: $-1 < x < 5$.

The answer is a whole stretch of the number line, not a list of numbers. Check one inside it: at x = 0 the factors are -5 and 1, and their product is negative, as asked. The roots themselves are out, since there the product is 0, not less than 0.