You are given a polynomial of degree two in $\displaystyle x$, $\displaystyle x^2-3x-10$. A convenient way of writing it is in the form $\displaystyle (x-x_1)(x-x_2)$ where $\displaystyle x_1$ and $\displaystyle x_2$ are its roots, this is, the numbers such that the polynomial is zero.
Now the reasoning: Because the polynomial is a continuous function, these two numbers $\displaystyle x_1,x_2$ set a change of sign in the expression $\displaystyle x^2-3x-10$. So you should check whether is positive or negative if $\displaystyle x$ is greater or smaller than $\displaystyle x_1$ and $\displaystyle x_2$.
Also for the absolute value condiction $\displaystyle |x-2|<a$ this means $\displaystyle -a<x-2<a$
Essentially, you are asked to solve an inequality: $\displaystyle x^2- 3x- 10< 0$. And the simplest way to do that is to solve the associated equation. As Ruun suggests, we can factor $\displaystyle x^2- 3x- 10= (x- 5)(x+ 2)= 0$. So where is that equal to 0? Those points separate "> 0" from "< 0". Just pick one value in each interval to see whether it is "> 0" or "< 0".