<s>[INST] $|x^2-5x+2|\leq 4$ 
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.
 [/INST] 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:
$$\left[\frac{5-\sqrt{33}}{2},2\right]\cup\left[3,\frac{5+\sqrt{33}}{2}\right]$$
</s>