We are asked to solve the equation: $\frac{2x-6}{x-1} - \frac{x+2}{x+1} = \frac{3x+4}{x^2+x-2}$.

1. Problem Description

We are asked to solve the equation:

$\frac{2x-6}{x-1} - \frac{x+2}{x+1} = \frac{3x+4}{x^2+x-2}$.

2. Solution Steps

First, we factor the denominator on the right side:

$x^2+x-2 = (x+2)(x-1)$.

The given equation can be written as:

$\frac{2x-6}{x-1} - \frac{x+2}{x+1} = \frac{3x+4}{(x+2)(x-1)}$

To eliminate the fractions, we multiply both sides of the equation by $(x-1)(x+1)(x+2)$,

but since $(x+2)(x-1) = x^2+x-2$ is in the denominator on the right side of the original equation, we multiply both sides of the equation by $(x-1)(x+1)(x+2) = (x+1)(x^2+x-2) = (x+2)(x^2-1)$:

$(2x-6)(x+1) - (x+2)(x-1) = 3x+4$

$2x^2 + 2x - 6x - 6 - (x^2 - x + 2x - 2) = 3x+4$

$2x^2 - 4x - 6 - (x^2 + x - 2) = 3x+4$

$2x^2 - 4x - 6 - x^2 - x + 2 = 3x+4$

$x^2 - 5x - 4 = 3x+4$

$x^2 - 5x - 3x - 4 - 4 = 0$

$x^2 - 8x - 8 = 0$

We use the quadratic formula to solve for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a=1$, $b=-8$, and $c=-8$.

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-8)}}{2(1)}$

$x = \frac{8 \pm \sqrt{64 + 32}}{2}$

$x = \frac{8 \pm \sqrt{96}}{2}$

$x = \frac{8 \pm \sqrt{16 \times 6}}{2}$

$x = \frac{8 \pm 4\sqrt{6}}{2}$

$x = 4 \pm 2\sqrt{6}$

The domain of the equation requires that $x \neq 1$, $x \neq -1$, and $x \neq -2$. Since $4+2\sqrt{6} \approx 8.9$ and $4-2\sqrt{6} \approx -0.9$, neither solution is extraneous.

3. Final Answer

$x = 4 \pm 2\sqrt{6}$