1. **State the problem:** Solve the quadratic equation $$2x^2 + 6x - 5 = 0$$ using the method of completing the square. 2. **Rewrite the equation:** Divide all terms by 2 to make the coefficient of $$x^2$$ equal to 1: $$x^2 + 3x - \frac{5}{2} = 0$$ 3. **Isolate the constant term:** Move the constant to the right side: $$x^2 + 3x = \frac{5}{2}$$ 4. **Complete the square:** Take half of the coefficient of $$x$$, which is $$\frac{3}{2}$$, and square it: $$\left(\frac{3}{2}\right)^2 = \frac{9}{4}$$ Add this to both sides: $$x^2 + 3x + \frac{9}{4} = \frac{5}{2} + \frac{9}{4}$$ 5. **Simplify the right side:** Convert $$\frac{5}{2}$$ to $$\frac{10}{4}$$ to add: $$\frac{10}{4} + \frac{9}{4} = \frac{19}{4}$$ 6. **Write the left side as a perfect square:** $$(x + \frac{3}{2})^2 = \frac{19}{4}$$ 7. **Take the square root of both sides:** $$x + \frac{3}{2} = \pm \sqrt{\frac{19}{4}} = \pm \frac{\sqrt{19}}{2}$$ 8. **Solve for $$x$$:** $$x = -\frac{3}{2} \pm \frac{\sqrt{19}}{2} = \frac{-3 \pm \sqrt{19}}{2}$$ **Final answer:** $$x = \frac{-3 + \sqrt{19}}{2} \quad \text{or} \quad x = \frac{-3 - \sqrt{19}}{2}$$