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.

You have two cases :
Case 1 : When $x^2-1\ge0$, you have $x^2-1\ge 2x-2.$
Case 2 : When $x^2-1\lt0$, you have $-(x^2-1)\ge 2x-2.$
The answer is $$"x^2-1\ge0\ \text{and}\ x^2-1\ge 2x-2"\ \text{or}\ "x^2-1\lt0\ \text{and}\ -(x^2-1)\ge 2x-2."$$
