1. **State the problem:** Solve the equation $$2(x + 1)(x - 4) - (x - 2)^2 = 0$$ for $x$. 2. **Expand and simplify:** $$2(x + 1)(x - 4) = 2(x^2 - 4x + x - 4) = 2(x^2 - 3x - 4) = 2x^2 - 6x - 8$$ $$(x - 2)^2 = x^2 - 4x + 4$$ 3. **Rewrite the equation:** $$2x^2 - 6x - 8 - (x^2 - 4x + 4) = 0$$ 4. **Simplify the left side:** $$2x^2 - 6x - 8 - x^2 + 4x - 4 = 0$$ $$x^2 - 2x - 12 = 0$$ 5. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=-2$, $c=-12$. 6. **Calculate discriminant:** $$b^2 - 4ac = (-2)^2 - 4(1)(-12) = 4 + 48 = 52$$ 7. **Write solutions:** $$x = \frac{-(-2) \pm \sqrt{52}}{2(1)} = \frac{2 \pm \sqrt{52}}{2}$$ 8. **Simplify square root:** $$\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}$$ 9. **Final solutions:** $$x = \frac{2 \pm 2\sqrt{13}}{2} = 1 \pm \sqrt{13}$$ **Answer:** $$x = 1 + \sqrt{13} \quad \text{or} \quad x = 1 - \sqrt{13}$$