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

 $x^2-|x+2|+x>0$
Case (i) :  $x+2\ge 0orx\ge -2$
Or   $x^2-x-2+x>0$
Or  $x^2-2>0$
 $\Rightarrow x\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$
$\Rightarrow x\in [-2,-\sqrt{2})\cup (\sqrt{2},\infty )$    $(\because x\ge -2)$(1)
Case (ii) :  $x+1<0orx<-2$
Or  $x^2+x+2+x>0$
Or  $x^2+2x+2>0$, which is true  $\forall x\in R$
Or   $x<-2$ (2)
From (1) and (2),  $x\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$