Let the function $$f(x) = 2x^3 + (2p-7)x^2 + 3(2p-9)x - 6$$ have a maxima for some value of $$x < 0$$ and a minima for some value of $$x > 0$$. Then, the set of all values of $$p$$ is

$$f(x) = 2x^3 + (2p - 7)x^2 + 3(2p - 9)x - 6$$

$$f'(x) = 6x^2 + 2(2p - 7)x + 3(2p - 9)$$

A local maxima for some value of $$x < 0$$ means a negative root

A local minima for some value of $$x > 0$$ means a positive root

$$6x^2 + 2(2p - 7)x + 3(2p - 9) = 0$$

$$\frac{C}{A} < 0 \implies \frac{3(2p - 9)}{6} < 0$$ (For roots to be of opposite signs)

$$2p < 9 \implies p < \frac{9}{2}$$

$$p \in \left(-\infty, \frac{9}{2}\right)$$