Answer
$\frac{1\pm i\sqrt {23} }{3}$
Work Step by Step
The quadratic formula says $x = \frac{ - b \pm \sqrt {b^2 - 4ac} }{2a}$ when $ax^2 + bx + c = 0$.
We know that $i^2=-1$.
If I multiply the equation by $2$ (this doesn't modify the solutions), the coefficients will be $9,-6,24$.
Hence here $x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot9\cdot24}}{2\cdot9}=\frac{6\pm\sqrt{36-864}}{18}=\frac{6\pm\sqrt{-828}}{18}=\frac{6\pm6i\sqrt {23}}{18}= \frac{1\pm i\sqrt {23} }{3}$
