smc

Plugging into the quadratic formula, we get $$x=\frac{2\sqrt{2}\pm \sqrt{8-4ac}}{2a}.$$ The discriminant is equal to 0, so this simplifies to $x=\frac{2\sqrt{2}}{2a}=\frac{\sqrt{2}}{a}.$ Because we are given that $a$ is real, $x$ is always real, and the answer is $\boxed{\text{equal and real}}.$