Absolute value quadratic inequalities not the usual? $ | -x^2 + 6x | \gt 13 $,for example. I would start off solving $ -x^2 + 6x = \pm 13 $ and either get 4 solutions, 3 solutions or two simply do the the nature of the graph. Without knowing if the two you get are the global minimums of the graph or if they are the two above the local maximum(the middle bump), how could one find out where $ y \gt 13 $?
It seems very hard to do this without a graphing calculator. Is this why I have never seen these types of problems in previous math classes? Due to their impracticality in the classroom and it's constraints(ie: time)?
 A: To start, notice that $|-x^2+6x| > 13$ iff $-x^2+6x > 13$ or $-x^2+6x < -13$. Hence, we need to solve these two inequalities and merge the two solutions. 
First inequality: 
$13 < -x^2+6x$ 
$x^2-6x+13 < 0$ 
$(x-3)^2+4 < 0$ (1)
Second inequality: 
$-13 > -x^2+6x$ 
$x^2-6x-13 > 0$ 
$(x-3)^2-22 > 0$ (2)
For what values of $x$ do inequalities (1) and (2) hold?
A: The Graphical Approach
You could start off with a bit of simplification $g(x) = |f(x)| = |-x^2 + 6x| = |x(x-6)|$.
Drawing the graph of $f(x) = x(x-6)$ shouldn't be particularly challenging. You know where the roots are (from inspection of the factorised form) and you can simply point-test to see whether each part of the graph ($x<0$, $0<x<6$ and $x>6$) lies above the x-axis or not.
After that, sketch the modulus curve. $|f(x)|$ means that all outputs must become positive, so simply reflect the bottom half of the curve along the x-axis and delete everything underneath it.
Then draw the line $y = 13$. Find the intersections between $g(x)$ and $y=13$ and simply look at the regions for which $g(x) > 13$.
A: You can distinguish between the following two cases:
Case 1: $-x^2+6x \geq 0$
First solve this.  Then the equation, which has to be solved is $-x^2+6x > 13$.
The two solutions I lable $L_{11}$ and $L_{12}$. Then the solution for the first case is  $L_1=L_{11} \wedge L_{12}$
Case 2: $-x^2+6x < 0$
Solve this.  Then the equation, which has to be solved is $x^2-6x > 13$.
Then the solution for the second case is  $L_2=L_{21} \wedge L_{22}$
The solution for the whole equation is $L = L_1 \vee L_2$
