1. **State the problem:** Solve the quartic equation $$x^4 - 5x^2 + 1 = 0$$. 2. **Use substitution:** Let $$y = x^2$$, so the equation becomes $$y^2 - 5y + 1 = 0$$. 3. **Solve the quadratic in $$y$$:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-5$$, and $$c=1$$. 4. Calculate the discriminant: $$\Delta = (-5)^2 - 4 \times 1 \times 1 = 25 - 4 = 21$$. 5. Find the roots for $$y$$: $$y = \frac{5 \pm \sqrt{21}}{2}$$. 6. **Back-substitute for $$x$$:** Since $$y = x^2$$, we have $$x^2 = \frac{5 + \sqrt{21}}{2}$$ or $$x^2 = \frac{5 - \sqrt{21}}{2}$$. 7. **Find $$x$$:** $$x = \pm \sqrt{\frac{5 + \sqrt{21}}{2}}$$ and $$x = \pm \sqrt{\frac{5 - \sqrt{21}}{2}}$$. 8. **Final answer:** The four real roots are $$x = \pm \sqrt{\frac{5 + \sqrt{21}}{2}}, \quad x = \pm \sqrt{\frac{5 - \sqrt{21}}{2}}$$.