The number of elements in the set $$\{x \in R : (|x| - 3)|x + 4| = 6\}$$ is equal to:

We solve $$(|x| - 3)|x + 4| = 6$$ by considering cases based on the sign of $$x$$ and $$x + 4$$.

Case 1: $$x \geq 0$$. Then $$|x| = x$$ and $$|x + 4| = x + 4$$, so $$(x - 3)(x + 4) = 6$$. Expanding: $$x^2 + x - 12 = 6$$, i.e., $$x^2 + x - 18 = 0$$. Using the quadratic formula: $$x = \frac{-1 \pm \sqrt{1 + 72}}{2} = \frac{-1 \pm \sqrt{73}}{2}$$. Since $$x \geq 0$$, only $$x = \frac{-1 + \sqrt{73}}{2} \approx 3.77$$ is valid. This gives one solution.

Case 2: $$-4 \leq x < 0$$. Then $$|x| = -x$$ and $$|x + 4| = x + 4$$, so $$(-x - 3)(x + 4) = 6$$. This gives $$-(x + 3)(x + 4) = 6$$, or $$x^2 + 7x + 12 = -6$$, i.e., $$x^2 + 7x + 18 = 0$$. The discriminant is $$49 - 72 = -23 < 0$$, so there are no real solutions in this case.

Case 3: $$x < -4$$. Then $$|x| = -x$$ and $$|x + 4| = -(x + 4)$$, so $$(-x - 3)(-(x + 4)) = 6$$, which simplifies to $$(x + 3)(x + 4) = 6$$. Expanding: $$x^2 + 7x + 12 = 6$$, i.e., $$x^2 + 7x + 6 = 0$$, giving $$(x + 1)(x + 6) = 0$$. Since $$x < -4$$, only $$x = -6$$ is valid. This gives one solution.

In total, there are 2 elements in the set.