Hence solve for \( x: \quad \frac{x^{2}-4 x}{3}-\frac{5}{x^{2}-4 x+2}=4 \)

First, let's simplify and rearrange the equation. Start with the original equation: \[ \frac{x^{2}-4x}{3} - \frac{5}{x^{2}-4x+2} = 4 \] First, we denote \( y = x^2 - 4x \). This transforms the equation to: \[ \frac{y}{3} - \frac{5}{y + 2} = 4 \] Next, we multiply both sides by \( 3(y + 2) \) to eliminate the denominators: \[ y(y + 2) - 15 = 12(y + 2) \] Expanding both sides gives: \[ y^2 + 2y - 15 = 12y + 24 \] Bringing all terms to one side results in: \[ y^2 - 10y - 39 = 0 \] Now we can solve this quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -10, c = -39 \): Calculating the discriminant: \[ b^2 - 4ac = (-10)^2 - 4 \cdot 1 \cdot (-39) = 100 + 156 = 256 \] Taking the square root: \[ \sqrt{256} = 16 \] Now substituting back into the quadratic formula: \[ y = \frac{10 \pm 16}{2} \] Calculating the two potential values for \( y \): 1. \( y = \frac{26}{2} = 13 \) 2. \( y = \frac{-6}{2} = -3 \) Now, substituting back to find \( x \). Recall that \( y = x^2 - 4x \): 1. For \( y = 13 \): \[ x^2 - 4x - 13 = 0 \] Using the quadratic formula again: \[ x = \frac{4 \pm \sqrt{16 + 52}}{2} = \frac{4 \pm \sqrt{68}}{2} = \frac{4 \pm 2\sqrt{17}}{2} = 2 \pm \sqrt{17} \] 2. For \( y = -3 \): \[ x^2 - 4x + 3 = 0 \] Again using the quadratic formula: \[ x = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2} \] Thus, \[ x = 3 \quad \text{or} \quad x = 1 \] So, the solutions for \( x \) are: \[ x = 2 + \sqrt{17}, \quad x = 2 - \sqrt{17}, \quad x = 3, \quad x = 1 \] Therefore, the final solutions to the equation are: \[ x = 2 + \sqrt{17}, \quad x = 2 - \sqrt{17}, \quad x = 3, \quad x = 1 \]