jmc

Let the roots be $r$ and $s$ (not necessarily real). We take the cases where $r=s$ and $r\ne s.$

Case 1: $r=s.$

Since $r$ is the only root, we must have $r^2-2=r.$ Then $r^2-r-2=0,$ which factors as $(r-2)(r+1)=0,$ so $r=2$ or $r=-1.$ This leads to the quadratics $x^2-4x+4$ and $x^2+2x+1.$

Case 2: $r\ne s.$

Each of $r^2-2$ and $s^2-2$ must be equal to $r$ or $s.$ We have three cases:

(i) $r^2-2=r$ and $s^2-2=s.$

(ii) $r^2-2=s$ and $s^2-2=r.$

(iii) $r^2-2=s^2-2=r$.

In case (i), as seen from Case $r,$ $s\in \{2,-1\}.$ This leads to the quadratic $(x-2)(x+1)=x^2-x-2.$

In case (ii), $r^2-2=s$ and $s^2-2=r.$ Subtracting these equations, we get $$r^2-s^2=s-r.$$Then $(r-s)(r+s)=s-r.$ Since $r-s\ne 0,$ we can divide both sides by $r-s,$ to get $r+s=-1.$ Adding the equations $r^2-2=s$ and $s^2-2=r,$ we get $$r^2+s^2-4=r+s=-1,$$so $r^2+s^2=3.$ Squaring the equation $r+s=-1,$ we get $r^2+2rs+s^2=1,$ so $2rs=-2,$ or $rs=-1.$ Thus, $r$ and $s$ are the roots of $x^2+x-1.$

In case (iii), $r^2-2=s^2-2=r.$ Then $r^2-r-2=0,$ so $r=2$ or $r=-1.$

If $r=2,$ then $s^2=4,$ so $s=-2.$ (We are assuming that $r\ne s.$) This leads to the quadratic $(x-2)(x+2)=x^2-4.$

If $r=-1$, then $s^2=1,$ so $s=1.$ This leads to the quadratic $(x+1)(x-1)=x^2-1.$

Thus, there are $\boxed{6}$ quadratic equations that work, namely $x^2-4x+4,$ $x^2+2x+1,$ $x^2-x-2,$ $x^2+x-1,$ $x^2-4,$ and $x^2-1.$