1. **State the problem:** Solve the equation $$20x^2 - 6 = 13ix\sqrt{3}$$ for $x$. 2. **Rewrite the equation:** We want to isolate $x$. The equation is $$20x^2 - 6 = 13ix\sqrt{3}$$. 3. **Bring all terms to one side:** $$20x^2 - 13ix\sqrt{3} - 6 = 0$$. 4. **Identify the type of equation:** This is a quadratic equation in $x$ with complex coefficients. 5. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Here, $$a = 20, \quad b = -13i\sqrt{3}, \quad c = -6$$. 6. **Calculate the discriminant:** $$b^2 - 4ac = (-13i\sqrt{3})^2 - 4 \times 20 \times (-6)$$ Calculate each part: $$(-13i\sqrt{3})^2 = (-13)^2 \times (i)^2 \times (\sqrt{3})^2 = 169 \times (-1) \times 3 = -507$$ $$-4ac = -4 \times 20 \times (-6) = 480$$ So, $$b^2 - 4ac = -507 + 480 = -27$$. 7. **Calculate the square root of the discriminant:** $$\sqrt{-27} = \sqrt{27}i = 3\sqrt{3}i$$. 8. **Apply the quadratic formula:** $$x = \frac{-(-13i\sqrt{3}) \pm 3\sqrt{3}i}{2 \times 20} = \frac{13i\sqrt{3} \pm 3\sqrt{3}i}{40}$$. 9. **Factor out $i\sqrt{3}$ in the numerator:** $$x = \frac{i\sqrt{3}(13 \pm 3)}{40}$$. 10. **Calculate the two solutions:** - For $+$ sign: $$x = \frac{i\sqrt{3} \times 16}{40} = \frac{16}{40} i\sqrt{3} = \frac{2}{5} i\sqrt{3}$$ - For $-$ sign: $$x = \frac{i\sqrt{3} \times 10}{40} = \frac{10}{40} i\sqrt{3} = \frac{1}{4} i\sqrt{3}$$ **Final answer:** $$x = \frac{2}{5} i\sqrt{3} \quad \text{or} \quad x = \frac{1}{4} i\sqrt{3}$$