Question 1209113
To solve this equation, we'll first clear the denominators by multiplying both sides by $(x+2)(x-4)(x+1)$:

$$2x(x-4)(x+1) = (x-3)(x+1)(x+2) + (2x+7)(x-4)(x+2)$$

Expand both sides:

$$2x^3 - 6x^2 - 8x = x^3 - 8x - 3 + 2x^3 - 2x^2 - 22x - 56$$

Combine like terms:

$$-x^2 - 12x - 59 = 0$$

This is a quadratic equation. We can solve it using the quadratic formula:

$$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(-1)(-59)}}{2(-1)}$$

Simplifying:

$$x = \frac{12 \pm \sqrt{-100}}{-2}$$

Since the discriminant is negative, there are no real solutions to this equation.