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

From the given data, $x^2 - |x + 2| + x > 0$

We can discuss this problem by observing two different cases.

Case 1: When $(x + 2) \geq 0$

Therefore, $x^2 - (x + 2) + x > 0$

Hence, $x^2 - 2 > 0$

$x^2 > 2$

$\Rightarrow x < -\sqrt{2}$ or $x > \sqrt{2}$

Hence, $x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$ ...... $(1)$

Case 2: When $(x + 2) < 0$

Then $x^2 + x + 2 + x > 0$

So, $x^2 + 2x + 2 > 0$

This gives $(x + 1)^2 + 1 > 0$ and this is true for every $x$ (Because $(x + 1)^2 \geq 0 \Rightarrow (x + 1)^2 + 1 \geq 1$)

Hence, $x \in (-\infty, \infty)$ ...... $(2)$

From equations $(1)$ and $(2)$, we take the intersection, hence we get $x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$.

Therefore, $x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$ is the required answer.