The coefficients $$a, b, c$$ in the quadratic equation $$ax^2 + bx + c = 0$$ are from the set $$\{1, 2, 3, 4, 5, 6\}$$. If the probability of this equation having one real root bigger than the other is $$p$$, then $$216p$$ equals :

$$D > 0$$.

$$D = b^2 - 4ac > 0 \implies b^2 > 4ac$$

Count pairs $$(a, c)$$ such that $$4ac < b^2$$ for each $$b \in \{1...6\}$$:

o $$b=1, 2$$: $$b^2 \le 4$$, no cases.

o $$b=3$$: $$b^2=9 > 4ac \implies ac < 2.25$$. Pairs: $$(1,1), (1,2), (2,1)$$ (3 cases)

o $$b=4$$: $$b^2=16 > 4ac \implies ac < 4$$. Pairs: $$(1,1), (1,2), (1,3), (2,1), (3,1)$$ (5 cases)

o $$b=5$$: $$b^2=25 > 4ac \implies ac < 6.25$$. Pairs: $$(1,1..6), (2,1..3), (3,1,2), (4,1), (5,1), (6,1)$$ (14 cases)

o $$b=6$$: $$b^2=36 > 4ac \implies ac < 9$$. Pairs: $$(1,1..6), (2,1..4), (3,1,2), (4,1,2), (5,1), (6,1), (8 \text{ is not possible})$$

Total for $$b=6$$ is 16 cases.

Total cases $$= 3 + 5 + 14 + 16 = 38$$.

Total sample space $$= 6^3 = 216$$.

$$p = 38/216$$. Therefore, $$216p = 38$$.