1. **State the problem:** Find the zeros of the quadratic equation using the quadratic formula. 2. **Quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a \neq 0$. 3. **Given values:** $$a = -1, \quad b = -8, \quad c = -13$$ 4. **Substitute values into the formula:** $$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot (-1) \cdot (-13)}}{2 \cdot (-1)}$$ 5. **Simplify inside the square root:** $$x = \frac{8 \pm \sqrt{64 - 52}}{-2}$$ 6. **Calculate the discriminant:** $$64 - 52 = 12$$ 7. **Rewrite the expression:** $$x = \frac{8 \pm \sqrt{12}}{-2}$$ 8. **Simplify $\sqrt{12}$:** $$\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$$ 9. **Rewrite the expression with simplified root:** $$x = \frac{8 \pm 2\sqrt{3}}{-2}$$ 10. **Divide numerator and denominator by 2 (showing cancellation):** $$x = \frac{\cancel{2}(4 \pm \sqrt{3})}{\cancel{2}(-1)} = \frac{4 \pm \sqrt{3}}{-1}$$ 11. **Simplify the division by $-1$:** $$x = -(4 \pm \sqrt{3})$$ 12. **Write the two solutions explicitly:** $$x_1 = -4 - \sqrt{3}$$ $$x_2 = -4 + \sqrt{3}$$ **Final answer:** $$x = -4 \pm \sqrt{3}$$