Find the domain of $\sqrt{x^2-9}$

Find the domain of $\sqrt{x^2-9}$

$$\sqrt{x^2-9}$$
I know that the domain of square root is greater than or equal to zero. I
solve for when $x^2-9<0$ and get $x^2<9$. Now I get $x<-3$ and $x<3$. I
know that the domain is $(-\infty,-3] \cup [3,\infty)$ The problem is I do
not understand how the $x<3$ gets flipped to $x>3$, am I doing this step
properly or is there another way to do it?