Solve the rational equation \(\frac{2x^2 - 3x - 2}{x^2 - x - 6} = \frac{4x - 3}{x + 1}\)

Step 1 ：Step 1: Factor the polynomials in the equation. We get \(\frac{2(x - 1)(x + 2)}{(x - 2)(x + 3)} = \frac{4(x - \frac{3}{4})}{x + 1}\)

Step 2 ：Step 2: Cross multiply the denominators, we get \(2(x - 1)(x + 2)(x + 1) = 4(x - \frac{3}{4})(x - 2)(x + 3)\)

Step 3 ：Step 3: Simplify the equation and set it to zero, we get \(2x^3 + x^2 - 6x - 4 = 4x^3 - \frac{3x^2}{2} - \frac{21x}{2} + 6\)

Step 4 ：Step 4: Collect like terms and simplify the equation, we get \(2x^3 - 4x^3 + x^2 + \frac{3x^2}{2} + 6x + \frac{21x}{2} - 4 - 6 = 0\)

Step 5 ：Step 5: Simplify further to get \(-2x^3 + \frac{5x^2}{2} + \frac{27x}{2} = 0\)

Step 6 ：Step 6: Factor the equation to find the roots, we get \(-2x(x - 3)(x + \frac{9}{2}) = 0\)

Step 7 ：Step 7: So, the solutions are \(x = 0, x = 3, x = -\frac{9}{2}\)