Answer:
(D) [tex]3 \pm i\sqrt{3}[/tex]
Step-by-step explanation:
We can solve this using the quadratic formula, where a is 1, b is -6, and c is 12.
[tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\frac{-(-6)\pm\sqrt{6^2-4\cdot1\cdot12}}{2\cdot1}\\\\\frac{6\pm\sqrt{36-48}}{2}\\\\\frac{6\pm\sqrt{-12}}{2}\\\\\frac{6\pm i\sqrt{12}}{2}\\\\3\pm i\sqrt{3}\\\\[/tex]
Hope this helped!
