The set of all real numbers x for which $$x^{2} - |x + 2 |+ x > 0$$, is

Case-1 $$(x -2)$$:

So, $$x+2>0$$ and so $$\left|x+2\right|=x+2$$

or, $$x^2-\left(x+2\right)+x>0$$

or, $$x^2-x-2+x>0$$

or, $$x^2-2>0$$

or, $$x>\sqrt{\ 2}$$ and $$x<-\sqrt{\ 2}$$

But here, $$x>-2$$

so, $$-2<x<-\sqrt{\ 2}$$ and $$x>\sqrt{\ 2}$$

Case-2 $$(x \leq-2)$$:

So, $$x \leq-2$$ and $$x+2 \leq 0$$

or, $$\left|x+2\right|=-\left(x+2\right)$$

or, $$x^2-\left(-\left(x+2\right)\right)+x>0$$

or, $$x^2+x+2+x>0$$

or, $$x^2+2x+2>0$$

or, $$\left(x+1\right)^2+1>0$$ and this is always true that $$\left(x+1\right)^2+1>0$$ for any value of $$x$$

So, for $$x<-2$$, the relation always holds true

So, combining case-1 and case-2, $$x$$ belongs to $$(-\infty, -\surd{2}) \cup (\surd{2}, \infty)$$