To find the set of all values of $ p \in \mathbb{R} $ for which both roots of the equation $ x^2-(p+2)x+(2p+9)=0 $ are negative real numbers, follow these steps:

Discriminant Condition:

The equation's discriminant $ D $ must be non-negative for real roots:

$ (p+2)^2 - 4(2p+9) \geq 0 $

Simplifying this:

$ p^2 + 4p + 4 - 8p - 36 \geq 0 \quad \Rightarrow \quad p^2 - 4p - 32 \geq 0 $

This can be factored as:

$ (p-8)(p+4) \geq 0 $

Meaning $ p \in (-\infty, -4] \cup [8, \infty) $. …… (1)

Sum of Roots Condition:

The sum of the roots (which is $ p+2 $) must be negative:

$ p + 2 < 0 \quad \Rightarrow \quad p < -2 $ …… (2)

Product of Roots Condition:

The product of the roots $ (2p + 9) $ must be positive:

$ 2p + 9 > 0 \quad \Rightarrow \quad p > -\frac{9}{2} $ …… (3)

Determine the Valid Interval:

Combine the results from conditions (1), (2), and (3). From conditions (1) and (2), we find $ p < -2 $:

Intersection of $ (-\infty, -4] $ and $ (-\frac{9}{2}, -2) $ gives:

$ p \in \left(-\frac{9}{2}, -4\right] $

Calculate $\beta - 2\alpha$:

With $\alpha = -\frac{9}{2}$ and $\beta = -4$, compute:

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

Therefore, the difference $\beta - 2\alpha$ is 5.