1. **State the problem:** Solve the equation $$\frac{x}{x-1} + \frac{x}{x+1} - \frac{3}{x-1} = 4$$ using the quadratic formula. 2. **Rewrite the equation:** Move all terms to one side to set the equation equal to zero: $$\frac{x}{x-1} + \frac{x}{x+1} - \frac{3}{x-1} - 4 = 0$$ 3. **Find a common denominator:** The denominators are $x-1$ and $x+1$. The common denominator is $(x-1)(x+1) = x^2 - 1$. 4. **Rewrite each term with the common denominator:** $$\frac{x(x+1)}{x^2 - 1} + \frac{x(x-1)}{x^2 - 1} - \frac{3(x+1)}{x^2 - 1} - \frac{4(x^2 - 1)}{x^2 - 1} = 0$$ 5. **Combine the numerators:** $$\frac{x(x+1) + x(x-1) - 3(x+1) - 4(x^2 - 1)}{x^2 - 1} = 0$$ 6. **Simplify the numerator:** $$x^2 + x + x^2 - x - 3x - 3 - 4x^2 + 4 = 0$$ 7. **Combine like terms:** $$x^2 + x^2 - 4x^2 + x - x - 3x - 3 + 4 = 0$$ $$(-2x^2) - 3x + 1 = 0$$ 8. **Multiply both sides by $-1$ to simplify:** $$\cancel{-}2x^2 - 3x + 1 = 0 \Rightarrow 2x^2 + 3x - 1 = 0$$ 9. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=3$, and $c=-1$. 10. **Calculate the discriminant:** $$b^2 - 4ac = 3^2 - 4(2)(-1) = 9 + 8 = 17$$ 11. **Write the solutions:** $$x = \frac{-3 \pm \sqrt{17}}{2 \times 2} = \frac{-3 \pm \sqrt{17}}{4}$$ 12. **Final answer:** $$x = \frac{-3 + \sqrt{17}}{4} \quad \text{or} \quad x = \frac{-3 - \sqrt{17}}{4}$$ These are the two solutions to the original equation, provided they do not make any denominator zero (check $x \neq 1$ and $x \neq -1$). Both solutions are valid since neither equals 1 or -1.