Find the set of all real number x for which x²-|x+2|+x>0?

Note that x=-2 is a solution.

If x>-2 the equation is equivalent to x² - (x+2) + x > 0, i.e.
x² - 2 > 0, with solutions all x in ((√2,∞) U (-∞,-√2]) ∩ (-2,∞) = (-2,-√2) U (√2,∞)

If x<-2 the equation is equivalent to x² + (x+2) + x > 0, i.e. x² + 2x + 2 > 0, i.e. (x+1)² + 1 > 0, which holds for all x since squares are never negative, but given the case hypotheses the solution in this case is (-∞,-2).

Putting it all together the solution is (-∞,-√2) U (√2,∞).

x < -1 - Sqrt[2] || x > 0

Short answer: |x| > √2
___
For x ≥ -2, the inequality becomes
.. x^2 -x -2 +x > 0
.. x^2 -2 > 0
.. (x -√2)*(x +√2) > 0 . . . true for x in [-2, -√2) U (√2, ∞)

For x ≤ -2, the inequality becomes
.. x^2 +x +2 +x > 0
.. (x +1)^2 +1 > 0 . . . true for all x ≤ -2

The union of these two solution sets is
.. x is in (-∞, -√2) U (√2, ∞)
or
.. |x| > √2