Answer
$(x-3)^2=-8y$, $(-1,-2),(7,-2)$.
See graph.
Work Step by Step
1. Given vertex $(3,0)$ and focus $(3,-2)$ (opens downward), we have $p=2$ and equation (as in a standard form) $(x-3)^2=-4(2)(y)$ or $(x-3)^2=-8y$
2. Let $y=-2$, we have $(x-3)^2=16$ and $x=3\pm4$, thus we have the two points that define the latus rectum $(-1,-2),(7,-2)$
3. See graph.