Candied quadratics

The probability the first sweet is orange is $\frac{6}{n}$ . The probability the second sweet is orange is $\frac{5}{n-1}$ , given that the first sweet was orange.

Together, we have the probability both sweets are orange

$P = \frac{6}{n} \frac{5}{n-1} = \frac{1}{3}$

Rearranging. we get

$6 \times 5 \times 3 = 90 = n(n-1) = n^{2}-n$

Thus,

$n^{2}-n-90 = \boxed{0}$