1. The problem is to solve everything, but since no specific problem is given, I will interpret this as solving a general algebraic equation example. 2. Let's solve the quadratic equation $$ax^2 + bx + c = 0$$ where $a \neq 0$. 3. The formula to find the roots of a quadratic equation is 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 $\Delta < 0$, there are two complex roots. 5. Example: Solve $$2x^2 - 4x - 6 = 0$$ 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. First root: $$x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3$$ 10. Second root: $$x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1$$ 11. Final answer: The solutions to the equation are $$x = 3$$ and $$x = -1$$.