jmc

Let $y=x^2,$ so $$y^2-2y-7=0.$$By the quadratic formula, the roots are $y=1\pm 2\sqrt{2},$ so $$x^2=1\pm 2\sqrt{2}.$$Since $1-2\sqrt{2}<0,$ we must have $x^2=1+2\sqrt{2},$ which has $\boxed{2}$ real roots.