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

$(-∞,-\sqrt{2}) ∪ (\sqrt{2},∞)$

In this case, we have

$|x+2|=x+2$

$∴x^2-|x+2|+x>0$

$⇒x^2 -(x+2)+x>0$

$⇒x^2-2>0$

$⇒ x<-\sqrt{2}$ or $x>\sqrt{2} ⇒ x∈ (-∞,-\sqrt{2}) ∪(\sqrt{2},∞)$

But, $x+2≥0$ i.e. $x ≥-2$

$∴x∈[-2,-\sqrt{2})∪(\sqrt{2},∞)$

CASE II When $x + 2 < 0$

$|x+2| =- (x+2)$

$∴x^2 -|x+2|+x>0$

$⇒x^2+x+2+x>0⇒ x^2+2x+2>0$

This is true for all real values of x as the discriminant of $x^2 + 2x + 2$ is less than zero.

So, $x+2<0$ i.e. $x∈ (-∞,-2)$ is the solution set.

Hence, the solution set is $[-2,-\sqrt{2})∪(\sqrt{2},∞)∪(-∞,-2)$ or, $(-∞,-\sqrt{2}) ∪ (\sqrt{2},∞)$