jmc

We will complete the square on the given quadratic expression to find the vertex. Factoring $-2$ from the first two terms, we have $$y=-2(x^2+6x)-15$$In order to make the expression inside the parentheses a perfect square, we need to add and subtract $(6/2{)}^2=9$ inside the parentheses. Doing this, we get $$y=-2(x^2+6x+9-9)-15\Rightarrow -2(x+3{)}^2+3$$The graph of an equation of the form $y=a(x-h{)}^2+k$ is a parabola with vertex at $(h,k)$, so the vertex of our parabola is at $(-3,3)$. Thus, $m+n=-3+3=\boxed{0}$.