1. **State the problem:** Solve the quartic equation $$z^4 - 20z^2 + 64 = 0$$ for $z$. 2. **Rewrite the equation:** Let $u = z^2$. Then the equation becomes a quadratic in $u$: $$u^2 - 20u + 64 = 0$$ 3. **Use the quadratic formula:** For $au^2 + bu + c = 0$, solutions are $$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=-20$, $c=64$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-20)^2 - 4 \times 1 \times 64 = 400 - 256 = 144$$ 5. **Find $u$ values:** $$u = \frac{20 \pm \sqrt{144}}{2} = \frac{20 \pm 12}{2}$$ 6. **Evaluate each root:** - For $+$ sign: $$u = \frac{20 + 12}{2} = \frac{32}{2} = 16$$ - For $-$ sign: $$u = \frac{20 - 12}{2} = \frac{8}{2} = 4$$ 7. **Recall $u = z^2$, so solve for $z$:** - When $z^2 = 16$, $$z = \pm 4$$ - When $z^2 = 4$, $$z = \pm 2$$ 8. **Final solution:** $$z = -4, -2, 2, 4$$ All roots are real numbers, no complex numbers needed here.