jmc

We begin by writing $-4z^2+20z-6$ as $-(4z^2-20z+6)$. We then complete the square for $4z^2-20z+6$.

We know that the binomial to be squared will be in terms of $2z+b$ because $(2z{)}^2=4z^2$. By expanding $(2z+b{)}^2$, we get $4z^2+4bz+b^2$. We get that $4bz=-20z$, so $b=-5$, which gives us $(2z-5{)}^2=4z^2-20z+25$.

Therefore, $-(4z^2-20z+6)=-(4z^2-20z+25-19)=-[(2z-5{)}^2-19]=-(2z-5{)}^2+19$.

Since $(2z-5{)}^2$ is at least zero because it is a square of a real number, $-(2z-5{)}^2$ is at most 0. Therefore, the maximum value of $-4z^2+20z-6$ is $\boxed{19}$.