Solve the inequality: $|x^2 − 4| < 2$ This is a question on a calculus assignment our class received, I am a little confused on a few parts to the solution, can someone clear a few things up with it?
Since $x^2-4 = 0$ that means $x = 2$ and $x = -2$ are turning points.
When $x < -2$ :
$x^2-4<2$
$x^2<6$
$x < \sqrt{6}$ and $x < - \sqrt{6}$
But the solution says that $x > -\sqrt{6}$, what am I not getting?

$$|x^2-4|<2 \\ \implies -2<x^2-4<2 \\ \implies 2<x^2<6 \\ \implies \sqrt{2}<\pm x<\sqrt{6} \\ \implies \sqrt{2}<x<\sqrt{6},\ \ -\sqrt{2}>x>-\sqrt{6}$$
