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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 $$(x<-2)$$:
So, $$x<-2$$ and $$x+2<0$$
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 $$(-\infty, -\surd{2}) \cup (\surd{2}, \infty)$$
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