<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] Hints: 


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

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

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

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