your equation doesn't look quite right.....
note that z - 2 real implies that z = (z - 2) + 2 is real, and vice versa.
so let u = z - 2, so we want the number of real roots to u12 = 6.
that is, we're looking for real factors of u12 - 6.
now u12 - 6 = (u6 + √6)(u6 - √6).
notice that u6 + √6 > 0 for any real u, so we don't get any solutions there. so any REAL roots of u12 - 6 must be real roots of u6 - √6.
again, we have:
each cubic can be factored again:
this gives us two real roots u (to get z we'll have to add 2 to u). what about the quadratic factors?
let's look at their discriminants:
b2 - 4c (the same for both quadratics, since they only differ in the sign for b).
so these have no real roots.