How many sides does a polygon have if it has 33 more diagonals than sides?
The number of diagonals of an $n$-sided polygon is $\frac{n(n-3)}{2}$, so according to the problem,

$$
\frac{n(n-3)}{2}=n+33
$$

or

$$
n^{2}-5 n-66=0
$$

from which the positive value of $n$ is 11.