1. **State the problem:** Solve the equation $$x^4 - 13x^2 + 40 = 0$$. 2. **Use substitution:** Let $$y = x^2$$, so the equation becomes $$y^2 - 13y + 40 = 0$$. 3. **Solve the quadratic equation:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-13$$, and $$c=40$$. 4. Calculate the discriminant: $$\Delta = (-13)^2 - 4 \times 1 \times 40 = 169 - 160 = 9$$. 5. Find the roots for $$y$$: $$y = \frac{13 \pm \sqrt{9}}{2} = \frac{13 \pm 3}{2}$$. 6. So, $$y_1 = \frac{13 + 3}{2} = \frac{16}{2} = 8$$ $$y_2 = \frac{13 - 3}{2} = \frac{10}{2} = 5$$. 7. **Back-substitute:** Recall $$y = x^2$$, so $$x^2 = 8$$ or $$x^2 = 5$$. 8. Solve for $$x$$: $$x = \pm \sqrt{8} = \pm 2\sqrt{2}$$ $$x = \pm \sqrt{5}$$. 9. **Final answer:** The solutions are $$x = \pm 2\sqrt{2}, \pm \sqrt{5}$$.