absolute value inequalities When answer this kind of inequality
$|2x^2-5x+2| < |x+1|$
I am testing the four combinations when both side are +, one is + and the other is - and the opposite and when they are both -.
When I check the negative options, I need to flip the inequality sign?
Thanks   

if you take $x<0$
$-(2x^2-5x+2)<-(x+1)$
$-2x^2+6x-1<0$
you gets roots:
$x_1=\frac {1}{2}(3−7\sqrt7)≈2.82$
$x_2=\frac {1}{2}(3+\sqrt7)≈0.18$
Becouse we have x<0  (x is negative and our roots are pozitive) there are solutions:
$x \in { \emptyset}$
if $X>0$
$2x^2 - 5x + 2 < x +1$
$2x^2 - 6x + 1 < 0$
you gets roots:
$ x_1 = \frac {1}{2}( 3-\sqrt 7) \approx 0.18 $
$ x_2 = \frac {1}{2}( 3+\sqrt7) \approx 2.82$
so you get 
$  \frac {1}{2}( 3-\sqrt7) \leq x \leq \frac {1}{2}( 3+\sqrt7) $
if we combine both results we get
$x \in  {( \frac {1}{2}( 3-\sqrt 7), \frac {1}{2}( 3+\sqrt 7) )}$
