Each of the two options will be "an inverse" of the function, but their corresponding domain and range would not be the same. For this one we want the inverse to have domain $x \geq 2\sqrt{2}$, and range $y \geq \sqrt{2}$.

One way to check which of the two we want to keep is to plug in the same value of $x$ for both and see  which one gives a value within the desired range. For example, plugging in $x = 3$ gives $$y = \dfrac{3 \pm \sqrt{3^2-8}}{2} = \dfrac{3\pm 1}{2},$$ which is equal to $2$ when we choose "$+$", and equal to $1$ when we choose "$-$". Thus, we want to keep the one with the "$+$".

If we want to be more precise, we can solve the inequality $$\dfrac{x\pm\sqrt{x^2-8}}{2} \geq \sqrt{2}.$$ This has solution $x = 2\sqrt{2}$ when we choose "$-$", and $x \geq 2\sqrt{2}$ when we choose "$+$", which is what we want.