jmc

The vertex form of a parabolic equation is $y=a(x-h{)}^2+k$. Since we are given that the vertex is at $(2,-4)$, we know that $h=2$ and $k=-4$. Plugging that into our equation gives $y=a(x-2{)}^2-4$. Now, substituting the other given point $(4,12)$ into the equation to solve for $a$, we have $$\begin{array}{rl}12& =a(4-2{)}^2-4\\ 16& =a(2{)}^2\\ 16& =4a\\ 4& =a\end{array}$$ So the equation for the graphed parabola is $y=4(x-2{)}^2-4$. The zeros of the quadratic occur when $y=0$, so plugging that value into the equation to solve for $x$, we have $0=4(x-2{)}^2-4\Rightarrow (x-2{)}^2=1$. Taking the square root of both sides yields $x-2=\pm 1$, so $x=3$ or $x=1$. Thus, $m=3$ and $n=1$, so $m-n=3-1=\boxed{2}$.