In solving quadratic inequalities, it must be the case that the right hand side of the equation is zero. So we have:
Then we find the critical numbers, these are values of $x$ that will make our inequality above zero.
We have $x=1$ and $x=-1$ as critical numbers.
My critical numbers then partitioned my real number line into 3 parts/subintervals. Namely:
In each part it is advisable to get a test value, a number that lies on the subintervals and substitute it on the LHS above and we must note the sign, this are the sign of the subintervals relative to our inequality.
If $x=-2$ we have positive sign so for all $x$ in the first subinterval we have $x^2-1$ is positive.
If $x=0$ we have a negative sign so for all $x$ in the second subinterval $x^2-1$ is negative.
Lastly if $x=2$ we have a positive sign, so for all $x$ in the third subinterval we have $x^2-1$ is positive.
Originally we have $x^2-1>0$ this means that $x^2-1$ is positive so our answer is $$(-\infty,-1)\cup (1,+\infty).$$
This is the general way to solve quadratic inequality.