The problem asks to solve the equation $\frac{6}{x-6} - \frac{2(x+14)}{x^2-4x-12} = \frac{4x+11}{x^2+3x+2}$.

1. Problem Description

The problem asks to solve the equation

$\frac{6}{x-6} - \frac{2(x+14)}{x^2-4x-12} = \frac{4x+11}{x^2+3x+2}$.

2. Solution Steps

First, we factor the denominators:

$x^2 - 4x - 12 = (x-6)(x+2)$

$x^2 + 3x + 2 = (x+1)(x+2)$

The given equation can be rewritten as:

$\frac{6}{x-6} - \frac{2(x+14)}{(x-6)(x+2)} = \frac{4x+11}{(x+1)(x+2)}$

Multiply both sides of the equation by the least common denominator (LCD) $(x-6)(x+2)(x+1)$:

$6(x+2)(x+1) - 2(x+14)(x+1) = (4x+11)(x-6)$

$6(x^2 + 3x + 2) - 2(x^2 + 15x + 14) = 4x^2 - 24x + 11x - 66$

$6x^2 + 18x + 12 - 2x^2 - 30x - 28 = 4x^2 - 13x - 66$

$4x^2 - 12x - 16 = 4x^2 - 13x - 66$

$-12x - 16 = -13x - 66$

$-12x + 13x = -66 + 16$

$x = -50$

Now, we need to check if this value makes any denominator equal to zero.

$x-6 = -50-6 = -56 \neq 0$

$x^2-4x-12 = (-50)^2 - 4(-50) - 12 = 2500 + 200 - 12 = 2688 \neq 0$

$x^2+3x+2 = (-50)^2 + 3(-50) + 2 = 2500 - 150 + 2 = 2352 \neq 0$

Thus, $x=-50$ is a valid solution.

3. Final Answer

$x = -50$