Let the set of all values of $$p \in \mathbb{R}$$, for which both the roots of the equation $$x^2 - (p+2)x + (2p+9) = 0$$ are negative real numbers, be the interval $$(\alpha, \beta]$$. Then $$\beta - 2\alpha$$ is equal to

For the equation $$x^2 - (p+2)x + (2p+9) = 0$$ to have both roots as negative real numbers, three conditions must hold:

Condition 1 (Discriminant $$\geq 0$$):

$$D = (p+2)^2 - 4(2p+9) = p^2 + 4p + 4 - 8p - 36 = p^2 - 4p - 32 = (p-8)(p+4) \geq 0$$

This gives $$p \leq -4$$ or $$p \geq 8$$.

Condition 2 (Sum of roots $$\lt 0$$):

Sum $$= p + 2 \lt 0$$, so $$p \lt -2$$.

Condition 3 (Product of roots $$\gt 0$$):

Product $$= 2p + 9 \gt 0$$, so $$p \gt -\frac{9}{2}$$.

Combining all conditions: $$p \gt -\frac{9}{2}$$, $$p \lt -2$$, and ($$p \leq -4$$ or $$p \geq 8$$).

The intersection is $$-\frac{9}{2} \lt p \leq -4$$, i.e., the interval $$\left(-\frac{9}{2}, -4\right]$$.

So $$\alpha = -\frac{9}{2}$$ and $$\beta = -4$$.

$$\beta - 2\alpha = -4 - 2\left(-\frac{9}{2}\right) = -4 + 9 = 5$$

Hence, the correct answer is Option C.