1. The problem is to find the zeros (roots) of a quadratic equation, which are the values of $x$ where the quadratic equals zero. 2. The general form of a quadratic equation is $$ax^2 + bx + c = 0$$ where $a$, $b$, and $c$ are constants and $a \neq 0$. 3. The zeros can be found using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ This formula gives two solutions corresponding to the plus and minus signs. 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 (not real). 5. To solve a specific quadratic, substitute the values of $a$, $b$, and $c$ into the formula and simplify step-by-step. 6. Example: For $2x^2 - 4x - 6 = 0$, $a=2$, $b=-4$, $c=-6$. 7. Calculate the discriminant: $$\Delta = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$ 8. Since $\Delta = 64 > 0$, there are two real roots. 9. Apply the quadratic formula: $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$ 10. Calculate each root: $$x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3$$ $$x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1$$ 11. Final answer: The zeros of the quadratic are $x=3$ and $x=-1$.