1. Let's create a new algebra question 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. Calculate the roots: $$x_1 = \frac{-(-4) + \sqrt{64}}{2 \times 2} = \frac{4 + 8}{4} = 3$$ $$x_2 = \frac{-(-4) - \sqrt{64}}{2 \times 2} = \frac{4 - 8}{4} = -1$$ 7. Final answer: The solutions are $$x=3$$ and $$x=-1$$.