More precise way of solving inequality I need to solve this function:
$$
\lvert x^2-1\rvert\ge 2x-2\\
$$
I solved this equation:  
For $x<0$, the solution is non existing, here I got negative root, when I tried to solve quadratic function and for $x\ge 0$ I got points $x_1=-1$ and $x_2=3$.
My question is:
How do I set the solution of equation. Is there any procedure, with wich I can determine is equation valid for $[-1,3]$ or $[-\infty, -1]\lor [3, +\infty]$.
I know that I can just set the numbers and see the result, but I just want to now is there any other different way to do this.
Thanks.

As $\displaystyle |x|=\begin{cases} x &\mbox{if }  x\ge0 \\-x  & \mbox{if } x<0 \end{cases} $
If $x^2-1\ge0\iff x\ge1$ or $x\le-1,$
we get $$x^2-1\ge2x-2\iff x^2-2x+1\ge0\iff (x-1)^2\ge0$$ which is true
If $x^2<1\iff -1<x<1\  \ \  \  (1),$
we get $$-(x^2-1)>2x-2\iff x^2+2x-3<0$$
$$\iff (x+3)(x-1)<0\iff -3<x<1\  \ \ \ (2)$$ 
Now using $(1),(2)-1<x<1$
So, $x$ can assume any real value
