**Case 1:** $x \ge -2$ In this case, $|x+4| = x+4$ and $|x+2| = x+2$. The equation becomes: $x(x+4) + 3(x+2) + 10 = 0$ $x^2 + 4x + 3x + 6 + 10 = 0$ $x^2 + 7x + 16 = 0$ The discriminant is $D = 7^2 - 4(1)(16) = 49 - 64 = -15 < 0$. So, there are no real solutions in this case.

**Case 2:** $-4 \le x < -2$ In this case, $|x+4| = x+4$ and $|x+2| = -(x+2)$. The equation becomes: $x(x+4) + 3(-(x+2)) + 10 = 0$ $x^2 + 4x - 3x - 6 + 10 = 0$ $x^2 + x + 4 = 0$ The discriminant is $D = 1^2 - 4(1)(4) = 1 - 16 = -15 < 0$. So, there are no real solutions in this case.

**Case 3:** $x < -4$ In this case, $|x+4| = -(x+4)$ and $|x+2| = -(x+2)$. The equation becomes: $x(-(x+4)) + 3(-(x+2)) + 10 = 0$ $-x^2 - 4x - 3x - 6 + 10 = 0$ $-x^2 - 7x + 4 = 0$ $x^2 + 7x - 4 = 0$ Using the quadratic formula: $x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-4)}}{2(1)} = \frac{-7 \pm \sqrt{49 + 16}}{2} = \frac{-7 \pm \sqrt{65}}{2}$ $x_1 = \frac{-7 + \sqrt{65}}{2} \approx \frac{-7 + 8.06}{2} \approx \frac{1.06}{2} \approx 0.53$ $x_2 = \frac{-7 - \sqrt{65}}{2} \approx \frac{-7 - 8.06}{2} \approx \frac{-15.06}{2} \approx -7.53$ Since we are considering $x < -4$, $x_1$ is not a solution, but $x_2 = \frac{-7 - \sqrt{65}}{2}$ is a solution because it is less than -4.

Therefore, there is only one real solution.