$|x^2-5x+2|\leq 4$ 
Solve the following inequality.
  $$|x^2-5x+2|\leq 4.$$

I know how to interpret $|x-a|$ as the distance between $x$ and $a$ along the $x$-axis, but how does one interpret an absolute value of $|ax^2+bx+c|$?
For the inequality $|x^2-5x+2|\leq 4$ I factored the LHS and got $$\left|\left(x+\frac{\sqrt{7}-5}{2}\right)\left(x+\frac{5-\sqrt{7}}{2}\right)\right|\leq 4.$$
I don't know how to case-divide the LHS.
 A: The absolute value is defined as $$|x|=\begin{cases} \hphantom{-}x,& x\geq 0\\ -x, & x < 0\end{cases}$$
so $$|x|\leq a$$ can be done by casework:


*

*$x \geq 0$


$x = |x| \leq a$, i.e. $0\leq x\leq a,$


*
*

*$x < 0$



$-x = |x| \leq a$, i.e. $-a\leq x< 0.$
Combined it gives us $$|x|\leq a \iff -a\leq x\leq a.$$
Now, by substitution we get $$|x^2-5x+2|\leq 4 \iff -4\leq x^2-5x+2 \leq 4$$ which boils down to solving two quadratic inequalities: $x^2-5x+6\geq 0$ and $x^2-5x-2\leq 0$ and taking intersection.
A: Hints: 


*

*$$|x^2-5x+2| \le 4 \iff -4 \le x^2-5x+2 \le4 \iff \frac{1}{4} \le x^2-5x+\frac{25}{4} \le \frac{33}{4}.$$

*$$x^2-5x +\frac{25}{4}=\left(x-\frac{5}{2}\right)^2.$$

*$$a \le y^2 \le b \iff  \sqrt{a} \le y \le \sqrt{b} \quad \text{or} \quad -\sqrt{b} \le y \le -\sqrt{a}.$$

A: It's just $$-4\leq x^2-5x+2\leq4,$$
which is $$x^2-5x-2\leq0$$ (which gives
$\frac{5-\sqrt{33}}{2}\leq x\leq\frac{5+\sqrt{33}}{2}$)
and $$x^2-5x+6\geq0$$ (which is
$x\geq3$ or $x\leq2$).
Finally, we obtain:
$$\left[\frac{5-\sqrt{33}}{2},2\right]\cup\left[3,\frac{5+\sqrt{33}}{2}\right]$$
A: One interpretation is $-4 \leq x^2-5x+2 \leq 4$. This means the set of $x$ values such that the inequality holds, is the domain. The graph here shows that pictorially. The graph is shaded wherever $x$ is such that the inequality holds.
