jmc

We can simplify the inequality to $z^2-40z+336\le 0$. We could solve for the roots using the quadratic formula, but there is an easier solution by factoring: $z^2-40z+336=(z-12)(z-28)$. Thus the parabola $z^2-40z+336$ changes sign at $z=12$ and $z=28$. The solution is either the interval $(-\infty ,12]\cup [28,\infty )$ or $[12,28]$. We test values to find that the quadratic is non-positive over the interval $\boxed{[12,28]}$.