Question

I have ti find the two solutions to this equation. This problem deals with imaginary numbers i

Answer 1

Answer:

$x = \dfrac{1}{6} + \dfrac{i\sqrt{47}}{6}$   or   $x = \dfrac{1}{6} - \dfrac{i\sqrt{47}}{6}$

Step-by-step explanation:

Use the quadratic formula.

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(4)}}{2(3)}$

$x = \dfrac{1 \pm \sqrt{1 - 48}}{6}$

$x = \dfrac{1 \pm \sqrt{-47}}{6}$

$x = \dfrac{1}{6} \pm \dfrac{i\sqrt{47}}{6}$

$x = \dfrac{1}{6} + \dfrac{i\sqrt{47}}{6}$   or   $x = \dfrac{1}{6} - \dfrac{i\sqrt{47}}{6}$