1. The problem is to solve a quadratic equation of the form $ax^2 + bx + c = 0$. 2. The formula to find the roots of a quadratic equation is the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ This formula gives the solutions for $x$ where the quadratic equals zero. 3. Important rules: - The term under the square root, $b^2 - 4ac$, is called the discriminant. - If the discriminant is positive, there are two real roots. - If it is zero, there is one real root (a repeated root). - If it is negative, the roots are complex. 4. To solve a specific quadratic equation, substitute the values of $a$, $b$, and $c$ into the formula. 5. For example, if the quadratic equation is $2x^2 - 4x - 6 = 0$, then: - $a = 2$, $b = -4$, $c = -6$ 6. Calculate the discriminant: $$\Delta = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$ 7. Since $\Delta = 64 > 0$, there are two real roots. 8. Apply the quadratic formula: $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$ 9. Calculate each root: - For the plus sign: $$x = \frac{4 + 8}{4} = \frac{12}{4} = 3$$ - For the minus sign: $$x = \frac{4 - 8}{4} = \frac{\cancel{4} - 8}{\cancel{4}} = \frac{-4}{4} = -1$$ 10. Therefore, the solutions to the quadratic equation $2x^2 - 4x - 6 = 0$ are: $$x = 3 \text{ and } x = -1$$