1. **State the problem:** Solve the quadratic equation $$2x^2 - 20x = x^2 - 19$$ by completing the square. 2. **Rewrite the equation:** Move all terms to one side: $$2x^2 - 20x - x^2 + 19 = 0$$ which simplifies to $$x^2 - 20x + 19 = 0$$ 3. **Complete the square:** The formula to complete the square for $$x^2 + bx$$ is $$\left(x - \frac{b}{2}\right)^2 = x^2 - bx + \left(\frac{b}{2}\right)^2$$ Here, $$b = 20$$, so $$\left(x - \frac{20}{2}\right)^2 = x^2 - 20x + 100$$ 4. Add and subtract 100 inside the equation to keep it balanced: $$x^2 - 20x + 19 = (x - 10)^2 - 100 + 19 = (x - 10)^2 - 81$$ 5. Set the equation equal to zero: $$(x - 10)^2 - 81 = 0$$ which gives $$(x - 10)^2 = 81$$ 6. **Solve for $$x$$:** Take the square root of both sides: $$x - 10 = \pm \sqrt{81}$$ $$x - 10 = \pm 9$$ 7. Finally, solve for $$x$$: $$x = 10 \pm 9$$ **Answer:** The solutions are $$x = 10 \pm 9$$, which corresponds to choice D.