#### Answer

There is no regular polygon with 8 diagonals.

#### Work Step by Step

We can find the number of vertices if a regular polygon has 8 diagonals:
$\frac{n(n-3)}{2} = 8$
$n(n-3) = 16$
$n^2-3n- 16 = 0$
We can use the quadratic formula to find the solutions of the equation:
$n = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$n = \frac{-(-3) \pm \sqrt{(-3)^2-(4)(1)(-16)}}{(2)(1)}$
$n = \frac{3 \pm \sqrt{9+64}}{2}$
$n = \frac{3 \pm \sqrt{73}}{2}$
$n = \frac{3 - \sqrt{73}}{2}~~$ or $~~n = \frac{3 + \sqrt{73}}{2}$
$n = -2.77~~$ or $~~n = 5.77$
Since $n$ is not a whole number, there is no regular polygon with 8 diagonals.