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

Case I: When $\mathrm{x}+2\ge 0 \mathrm{i}.\mathrm{e}. \mathrm{x}\ge -2,$

Then given inequality becomes

${\mathrm{x}}^2-(\mathrm{x}+2)+\mathrm{x}>0\Rightarrow {\mathrm{x}}^2-2>0\Rightarrow \left|\mathrm{x}\right|>\sqrt{2}$

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

As $\mathrm{x}\ge -2,$ therefore, in this case the part of the solution set is $[-2, -\sqrt{2}]\cup (\sqrt{2}, \infty ).$

Case II: When $\mathrm{x}+2\le 0 \mathrm{i}.\mathrm{e}., \mathrm{x}\le -2,$

Then given inequality becomes ${\mathrm{x}}^2+(\mathrm{x}+2)+\mathrm{x}>0$

$\Rightarrow {\mathrm{x}}^2+2\mathrm{x}+2>0\Rightarrow {(\mathrm{x}+1)}^2+1>0,$ which is true for all real x

Hence, the part of the solution set in this case is $(-\infty , -2].$ Combining the two cases, the solution set is 

$(-\infty ,-2)\cup \left(\right[-2,-\sqrt{2}]\cup (\sqrt{2},\infty )=(-\infty , -\sqrt{2})\cup (\sqrt{2},\infty ).$