1. **State the problem:** Solve the quadratic equation $2x^2 - 3x - 6 = 0$ by completing the square. 2. **Rewrite the equation:** Divide all terms by 2 to make the coefficient of $x^2$ equal to 1: $$x^2 - \frac{3}{2}x - 3 = 0$$ 3. **Isolate the constant term:** $$x^2 - \frac{3}{2}x = 3$$ 4. **Complete the square:** Take half of the coefficient of $x$, which is $-\frac{3}{2}$, half is $-\frac{3}{4}$, then square it: $$\left(-\frac{3}{4}\right)^2 = \frac{9}{16}$$ Add this to both sides: $$x^2 - \frac{3}{2}x + \frac{9}{16} = 3 + \frac{9}{16}$$ 5. **Rewrite the left side as a perfect square:** $$\left(x - \frac{3}{4}\right)^2 = \frac{48}{16} + \frac{9}{16} = \frac{57}{16}$$ 6. **Take the square root of both sides:** $$x - \frac{3}{4} = \pm \sqrt{\frac{57}{16}} = \pm \frac{\sqrt{57}}{4}$$ 7. **Solve for $x$:** $$x = \frac{3}{4} \pm \frac{\sqrt{57}}{4} = \frac{3 \pm \sqrt{57}}{4}$$ **Final answer:** $$x = \frac{3 + \sqrt{57}}{4} \quad \text{or} \quad x = \frac{3 - \sqrt{57}}{4}$$