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   

Let me throw in my 5 cent. This type of inequality is of a very special type, the solution of which follows from the fact that you are allowed to square both sides, since the absolute value can't be negative. Therefore 
$$
\left(\left|2x^2-5x+2\right|\right)^2<\left(\left|x+1\right|\right)^2
$$
Here the absolute values can be simply dropped. Doing so and applying the following formula
$$
a^2-b^2=(a-b)(a+b)
$$
yields 
$$
\left( 2x^2-6x+1 \right)\left(2x^2-4x+3 \right)<0.
$$
The second term $2x^2-4x+3 $ represents a parabola, which never crosses the x-axis, since the discriminant is negative. Therefore all you need is to check for which values of $x$ the first term gets negative. Factoring the first quadratic equation yields
$$
\left( x- \frac{3-\sqrt{7}}{2} \right)\left(x- \frac{3+\sqrt{7}}{2}\right)\left(\text{Term} >0: \forall x \in \mathbb{R}\right)<0
$$
Therefore the solution is 
$$
x\in\left(\frac{3-\sqrt{7}}{2},\frac{3+\sqrt{7}}{2}\right)
$$
P.S.
The formula
$$
g(x)<|f(x)|<-g(x)
$$
works quick and fine mainly in cases where a single absolute value is present one side of the inequality. Say if we had
$$
\left|2x^2-5x+2\right|<x+1
$$
