how to solve this type equation

$$x^2+2\sqrt{2}x+2 = 0$$

$$\implies (x+\sqrt{2})^2 = 0$$

$$\implies x = -\sqrt{2},-\sqrt{2}$$

$$y^2-2 = 0$$

$$\implies y=-\sqrt{2},\sqrt{2}$$

Therefore , $$x \le y$$