Inequality with absolute value with unknowns

Question

I am preparing for a test and I do not understand how to get to the answer as written in the question.

Question: $x^2+4x-8<|x+2|$

Answer: $(−6;2)$

I can solve "$x^2+4x-8<|x+2|$" with the answer being "$x<2$" but I am having problem with the negative of the absolute value.

Answer 1

You have $x^2+4x-8<|x+2|$.

Case 1: $x+2\geq0$.

Then $|x+2|=x+2$ and $$x^2+4x-8<x+2 \iff x^2+3x-10<0 \iff (x-2)(x+5)<0 \\\iff x\in(-5,2)\cap [-2,+\infty)=[-2,2)=S_1$$

Case 2: $x+2<0$.

Then $|x+2|=-(x+2)$ and $$x^2+4x-8<-(x+2) \iff x^2+5x-6<0 \iff (x-1)(x+6)<0 \\\iff x\in(-6,1)\cap(-\infty,-2)=(-6,-2)=S_2$$

Finally, the solution is $S_1\cup S_2=(-6,2)$.

Answer 2

\begin{align} x^2+4x-8<|x+2| &\iff (x+2)^2-12 < |x+2| \\ &\iff |x+2|^2-|x+2|-12 < 0 \\ &\iff 4|x+2|^2-4|x+2|+1 < 49 \\ &\iff (2|x+2|-1)^2 < 49 \\ &\iff -7 < 2|x+2|-1 < 7 \\ &\implies |x+2| < 4 \\ &\implies -6 < x < 2 \end{align}