1. Let's create a practice algebra problem involving solving a quadratic equation. 2. Problem: Solve the quadratic equation $$2x^2 - 4x - 6 = 0$$ for $x$. 3. Formula: To solve a quadratic equation of the form $$ax^2 + bx + c = 0$$, we use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 4. Important rules: - The discriminant $$\Delta = b^2 - 4ac$$ determines the nature of the roots. - If $$\Delta > 0$$, there are two distinct real roots. - If $$\Delta = 0$$, there is one real root (a repeated root). - If $$\Delta < 0$$, the roots are complex. 5. Identify coefficients: $$a=2$$, $$b=-4$$, $$c=-6$$. 6. Calculate the discriminant: $$\Delta = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$ 7. Since $$\Delta = 64 > 0$$, there are two distinct real roots. 8. Calculate the roots: $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$ 9. Find each root: - $$x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3$$ - $$x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1$$ 10. Final answer: The solutions to the equation $$2x^2 - 4x - 6 = 0$$ are $$x = 3$$ and $$x = -1$$.