1. **Problem Statement:** Find the zeros of a quadratic function, which are the values of $x$ where the function equals zero. 2. **Formula:** The quadratic function is generally written as $f(x) = ax^2 + bx + c$. 3. **Zeros of the function:** To find zeros, solve $ax^2 + bx + c = 0$. 4. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ This formula gives the roots of the quadratic equation. 5. **Discriminant:** The term under the square root, $\Delta = b^2 - 4ac$, determines the nature of the roots: - If $\Delta > 0$, two distinct real roots. - If $\Delta = 0$, one real root (a repeated root). - If $\Delta < 0$, no real roots (complex roots). 6. **Example:** For $f(x) = 2x^2 - 4x - 6$, identify $a=2$, $b=-4$, $c=-6$. 7. Calculate discriminant: $$\Delta = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$ 8. Since $\Delta = 64 > 0$, two real roots exist. 9. Calculate roots: $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$ 10. Roots are: $$x_1 = \frac{4 + 8}{4} = 3$$ $$x_2 = \frac{4 - 8}{4} = -1$$ **Final answer:** The zeros of the quadratic function are $x=3$ and $x=-1$.