jmc

Note that $(x-7)(x+5)=x^2-2x-35$ and $(x-3)(x+1)=x^2-2x-3,$ so $$(x^2-2x-35)(x^2-2x-3)=1680.$$Let $y=x^2-2x-19,$ so $$(y-16)(y+16)=1680.$$Then $y^2-256=1680,$ so $y^2=1936.$ Hence, $y=\pm 44.$

If $y=44,$ then $x^2-2x-19=44,$ or $x^2-2x-63=0.$ The roots are $x=9$ and $x=-7.$

If $y=-44$, then $x^2-2x-19=-44,$ or $x^2-2x+25=0.$ This quadratic has no real roots.

Thus, the real roots are $\boxed{9,-7}.$