1. **State the problem:** Solve the equation $$9x^4 + 26x^2 = 3$$ for $x$. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$9x^4 + 26x^2 - 3 = 0$$ 3. **Substitution:** Let $y = x^2$. Then the equation becomes a quadratic in $y$: $$9y^2 + 26y - 3 = 0$$ 4. **Use the quadratic formula:** For $ay^2 + by + c = 0$, the solutions are $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=9$, $b=26$, and $c=-3$. 5. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 26^2 - 4 \times 9 \times (-3) = 676 + 108 = 784$$ 6. **Find the roots for $y$:** $$y = \frac{-26 \pm \sqrt{784}}{18} = \frac{-26 \pm 28}{18}$$ 7. **Calculate each root:** - For $+$: $$y = \frac{-26 + 28}{18} = \frac{2}{18} = \frac{1}{9}$$ - For $-$: $$y = \frac{-26 - 28}{18} = \frac{-54}{18} = -3$$ 8. **Recall $y = x^2$:** - If $x^2 = \frac{1}{9}$, then $x = \pm \frac{1}{3}$. - If $x^2 = -3$, then $x = \pm \sqrt{-3} = \pm i\sqrt{3}$. 9. **Final answer:** $$x = \pm \frac{1}{3}, \pm i\sqrt{3}$$