Please help with absolute value $|x^2 - 3x| = 28$ Just a question about solving an absolute value equation:
$$|x^2 - 3x| = 28$$
Do I just solve this as if the absolute value brackets weren't even there?
$$x^2 - 3x - 28 = 0$$
$$(x+4)(x-7) = 0$$
So $x=-4$ ; $x=7$
But I'm still confused why the absolute value signs would be there in the first place :(
EDIT:
So, I've found that $x=7, x=4, x=-4$
Just not $100\%$ now if they are correct as I've had a look at a few online abs value calculators to check my answers and only $x=-4$ and $x=7$ come up as answers.
Am I correct?
EDIT 2
Ok, $x=4$ can't work out. I found it by:
\begin{align*}
x^2 - 3x & = 28\\
x^2 - 3x - 28 & = 0\\
(x-7)(x-4) & = 0
\end{align*}
My answers here are $7$ and $4$.
So I'm lost as to why I got that answer! :(

Hint: you have to break it up into two cases - case 1: $x^2-3x\ge0$ and case 2: $x^2-3x<0$.
In case 1, your equation becomes $x^2-3x=28$ and in case 2, your equation becomes $3x-x^2=28$.
