1. **State the problem:** Solve the quadratic equation $$2x^2 - x = 6$$ for $x$. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$2x^2 - x - 6 = 0$$ 3. **Identify coefficients:** Here, $a = 2$, $b = -1$, and $c = -6$. 4. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 5. **Calculate the discriminant:** $$b^2 - 4ac = (-1)^2 - 4 \times 2 \times (-6) = 1 + 48 = 49$$ 6. **Substitute values into the formula:** $$x = \frac{-(-1) \pm \sqrt{49}}{2 \times 2} = \frac{1 \pm 7}{4}$$ 7. **Find the two solutions:** - For the plus sign: $$x = \frac{1 + 7}{4} = \frac{8}{4} = 2$$ - For the minus sign: $$x = \frac{1 - 7}{4} = \frac{-6}{4} = \frac{\cancel{6}}{\cancel{4}} \times \frac{-1}{1} = -\frac{3}{2}$$ 8. **Final answer:** $$x = 2 \quad \text{or} \quad x = -\frac{3}{2}$$