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?

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 $x=-10$.
