Solve the equation. \[ \frac{2 x}{x-2}=5+\frac{13 x^{2}}{x-2} \]

To solve the equation: \[ \frac{2x}{x-2} = 5 + \frac{13x^2}{x-2} \] **Step 1: Combine the Fractions** First, subtract \(5\) from both sides to group the terms with the common denominator: \[ \frac{2x}{x-2} - 5 = \frac{13x^2}{x-2} \] To combine the fractions, express \(5\) with the common denominator \(x - 2\): \[ \frac{2x}{x-2} - \frac{5(x-2)}{x-2} = \frac{13x^2}{x-2} \] Now, combine the numerators: \[ \frac{2x - 5(x - 2)}{x-2} = \frac{13x^2}{x-2} \] Simplify the numerator: \[ 2x - 5x + 10 = -3x + 10 \] So the equation becomes: \[ \frac{-3x + 10}{x-2} = \frac{13x^2}{x-2} \] **Step 2: Eliminate the Denominator** Since \(x \neq 2\) (to avoid division by zero), multiply both sides by \(x - 2\): \[ -3x + 10 = 13x^2 \] **Step 3: Rearrange into Standard Quadratic Form** Bring all terms to one side: \[ 13x^2 + 3x - 10 = 0 \] **Step 4: Solve the Quadratic Equation** Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 13\), \(b = 3\), and \(c = -10\): \[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 13 \cdot (-10)}}{2 \cdot 13} = \frac{-3 \pm \sqrt{529}}{26} = \frac{-3 \pm 23}{26} \] This gives two solutions: \[ x = \frac{20}{26} = \frac{10}{13} \quad \text{and} \quad x = \frac{-26}{26} = -1 \] **Step 5: Verify the Solutions** Ensure that neither solution makes the denominator \(x - 2\) zero. Both \(x = \frac{10}{13}\) and \(x = -1\) satisfy this condition. **Final Answer:** The solutions are \( x = \dfrac{10}{13} \) and \( x = -1 \).