1. Let's create an algebra problem involving quadratic equations. 2. Problem: Solve the quadratic equation $$2x^2 - 4x - 6 = 0$$. 3. Formula: Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-4$, and $c=-6$. 4. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$. 5. Since $$\Delta > 0$$, there are two real solutions. 6. Compute the roots: $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$. 7. First root: $$x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3$$. 8. Second root: $$x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1$$. 9. Final answer: The solutions are $$x = 3$$ and $$x = -1$$.