jmc

Given the number of sides of a polygon, $n$, the number of diagonals is given by $D=\frac{n(n-3)}{2}$. To find the number of sides given the number of diagonals, we can solve this equation for $n$. $$\begin{array}{rl}D& =\frac{n(n-3)}{2}\\ 2D& =n^2-3n\\ 0& =n^2-3n-2D.\end{array}$$ Then, using the quadratic formula we have $n=\frac{3\pm \sqrt{3^2-4(1)(-2D)}}{2(1)}=\frac{3\pm \sqrt{9+8D}}{2}$.

Since we are given that $D=9$, we have $n=\frac{3\pm \sqrt{9+8(9)}}{2}=\frac{3\pm 9}{2}=-3\text{ or }6$. Since we must have a positive number of sides, a polygon with nine diagonals has $\boxed{6}$ sides.