Answer: 150

Small Hint:

Extend $\overline{AD}$ and $\overline{BC}$ to meet at $P;$ the two $60^\circ$ angles make triangle $ABP$ equilateral

Big Hint:

Apply the Law of Cosines in triangle $PDC,$ where $PD = AB - 10,$ $PC = AB - 8,$ and $\angle P = 60^\circ$

Solution:

Extend rays $AD$ and $BC$ until they meet at $P.$ Triangle $ABP$ has $60^\circ$ angles at $A$ and $B,$ so it is equilateral: $PA = PB = AB.$ Writing $x = AB,$ we get $PD = PA - AD = x - 10$ and $PC = PB - BC = x - 8.$

The Law of Cosines in triangle $PDC,$ with $\angle P = 60^\circ$ and $DC = 12,$ gives $$\begin{aligned} 144 &= (x-10)^2 + (x-8)^2 \\ &\quad {}- (x-10)(x-8) \\ &= x^2 - 18x + 84, \end{aligned}$$ so $x^2 - 18x - 60 = 0$ and $x = 9 + \sqrt{81 + 60} = 9 + \sqrt{141}.$

Thus $p + q = 9 + 141 = 150.$