1. **State the problem:** We need to find the number of solutions to the equation $$x^4 = 3 - 20x^2$$. 2. **Rewrite the equation:** Bring all terms to one side: $$x^4 + 20x^2 - 3 = 0$$ 3. **Substitute:** Let $$t = x^2$$, then the equation becomes: $$t^2 + 20t - 3 = 0$$ 4. **Solve the quadratic in $$t$$:** Use the quadratic formula: $$t = \frac{-20 \pm \sqrt{20^2 - 4 \times 1 \times (-3)}}{2} = \frac{-20 \pm \sqrt{400 + 12}}{2} = \frac{-20 \pm \sqrt{412}}{2}$$ 5. **Simplify the square root:** $$\sqrt{412} = \sqrt{4 \times 103} = 2\sqrt{103}$$ 6. **Calculate the roots:** $$t = \frac{-20 \pm 2\sqrt{103}}{2} = -10 \pm \sqrt{103}$$ 7. **Evaluate the roots:** - $$t_1 = -10 + \sqrt{103} \approx -10 + 10.1489 = 0.1489$$ (positive) - $$t_2 = -10 - \sqrt{103} \approx -10 - 10.1489 = -20.1489$$ (negative) 8. **Interpret the roots:** Since $$t = x^2$$, and $$x^2$$ cannot be negative, discard $$t_2$$. 9. **Find $$x$$ values:** For $$t_1 = 0.1489$$, $$x^2 = 0.1489 \implies x = \pm \sqrt{0.1489} \approx \pm 0.3858$$ 10. **Number of solutions:** There are 2 real solutions to the original equation corresponding to $$x = \pm 0.3858$$. **Final answer:** The equation $$x^4 = 3 - 20x^2$$ has 2 real solutions.