Solve the equation : $x^2 − 6 |x − 2| − 28 = 0$ The following is an absolute value quadratic equation that I want to solve:
 $$x^2 − 6 |x − 2| − 28 = 0$$
Here is what I did : 
$x^2 − 6 |x − 2| − 28 = 0$
$x^2 − 6 |x − 2| − 28 = 0$
$-6|x-2|=28-x^2$
$6|x-2|=x^2-28$
$6x-12=x^2-28$  or  $28-x^2$
(Is this step correct ?)
Solving this two quadratic equations I get the answers $x=-2,8,-10,4$
But when I actually substitute these in the original equation I see that only $x=8,-10  $ satisfy and other two solutions are invalid . Why so?
 A: I think if you consider $x\ge 2$ and $x<2$ separately, then you have the following equations respectively:
$$x^2-6(x-2)-28=0\to x^2-6x-16=0$$
$$x^2+6(x-2)-28=0\to x^2+6x-40=0$$


A: HINT:
First of all, let $x=a+ib$ where $a,b$ are real
So, we have $$(a+ib)^2-28=6|a+ib-2|\implies a^2-b^2-28+2ab i=6\sqrt{(a-2)^2+b^2}$$
Equating the imaginary parts, we have $ab=0$
If $a=0,-(b^2+28)=6\sqrt{4+b^2}>0$ which is impossible
So, $b$ must be $0\implies x$ must be real
We know for real $m,$ $$|m|=\begin{cases} +m &\mbox{if } m\ge0 \\
-m & \mbox{if } m<0 \end{cases} $$
$$\text{So, }|x-2|=x-2\text{ if } x-2\ge0 \iff x\ge2$$
In that case, we have $$x^2-6(x-2)-28=0\implies x^2-6x-16=0\implies x=8\text{ or } -2$$
But we have $x\ge2,$ so we need to discard the solution $x=-2$
Similarly for $|x-2|=-(x-2)$ if $x-2<0\iff x<2$ 
A: Once you know that $6|x-2| = x^2-28$, you know that $x^2 - 28$ must be non-negative. This accounts for why $-2$ and $4$ are not solutions.
A: Once you get to $6|x-2|=x^2-28$, you need to consider two cases. 
Case $x \ge 2:$
If $x \ge 2$, then $6x-12 = x^2-28 \implies x^2-6x-16=0 \implies (x+2)(x-8)=0
\implies x \in \{-2, 8\}$ But we assumed $x \ge 2$, so we have to reject $x=-2$ as an extraneous root. Hence if $x \ge 2$, then $x=8$.
Case $x < 2:$
If $x < 2$, then $12-6x = x^2-28 \implies x^2+6x-40=0 \implies (x-4)(x+10)=0
\implies x \in \{4, -10\}$ But we assumed $x < 2$, so we have to reject $x=4$ as an extraneous root. Hence if $x < 2$, then $x=-10$.
