Absolute value inequality, where am I wrong? Tried to solving $|x^2-5x+5|<1$ using the square method, but I don't know what I did wrong:
$$-1<x^2-5x+5<1$$
$$-6<x^2-5x<-4$$
$$-6+\frac{25}{4}<x^2-5x+\frac{25}{4}<-4+\frac{25}{4}$$
$$\frac{25-24}{4}<\left(x-\frac{5}{2}\right)^2<\frac{25-16}{4}$$
$$\frac{\pm\sqrt1}{\sqrt4}<\sqrt{\left(x-\frac{5}{2}\right)^2}<\frac{\pm\sqrt9}{\sqrt4}$$
$$\frac{\pm1}{2}<x-\frac{5}{2}<\frac{\pm3}{2}$$
$$\frac{5\pm1}{2}<x<\frac{5\pm3}{2}$$
Possible solutions:
$2<x<1$ (not valid)
$2<x<4$ (ok)
$3<x<1$ (not valid)
$3<x<4$ (ok, but is a subset of solution 2.
Therefore $S=\{2<x<4\}$
The only problem is that the correct solution is $S=\{1<x<2\text{ or }3<x<4\}$. Where am I wrong?

You have a $\pm$ on each side of the inequality, but you need to change the direction of inequality for the "minus".
So you would have $$\dfrac{5 + 1}{2} < x < \dfrac{5+3}{2}$$
(The $+$'s go together), or $$\dfrac{5 - 1}{2} > x > \dfrac{5 - 3}{2}$$
(The $-$'s go together)
