We need to consider three cases based on the value of $x$:

Case 1: $x \le -4$
In this case, $|x| = -x$ and $|x+4| = -(x+4)$. The equation becomes:
$(-x - 3)(-x - 4) = 6$
$(x + 3)(x + 4) = 6$
$x^2 + 7x + 12 = 6$
$x^2 + 7x + 6 = 0$
$(x + 1)(x + 6) = 0$
$x = -1$ or $x = -6$
Since $x \le -4$, we only accept $x = -6$.

Case 2: $-4 < x < 0$
In this case, $|x| = -x$ and $|x+4| = x+4$. The equation becomes:
$(-x - 3)(x + 4) = 6$
$-x^2 - 4x - 3x - 12 = 6$
$-x^2 - 7x - 18 = 0$
$x^2 + 7x + 18 = 0$
The discriminant is $D = 7^2 - 4(1)(18) = 49 - 72 = -23 < 0$. So, there are no real solutions in this case.

Case 3: $x \ge 0$
In this case, $|x| = x$ and $|x+4| = x+4$. The equation becomes:
$(x - 3)(x + 4) = 6$
$x^2 + 4x - 3x - 12 = 6$
$x^2 + x - 18 = 0$
Using the quadratic formula, $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-18)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 72}}{2} = \frac{-1 \pm \sqrt{73}}{2}$
Since $x \ge 0$, we take the positive root: $x = \frac{-1 + \sqrt{73}}{2} \approx \frac{-1 + 8.54}{2} \approx 3.77$.
So, $x = \frac{-1 + \sqrt{73}}{2}$ is a valid solution.

Therefore, the solutions are $x = -6$ and $x = \frac{-1 + \sqrt{73}}{2}$. The number of elements in the set is 2.