Answer
$x_{1}=\dfrac{1 + \sqrt{31}i}{2}$ and $x_{2}=\dfrac{1- \sqrt{31}i}{2}$
Work Step by Step
Given $m^2-m+8=0$
$a=1, \ b=-1, \ c=8$
Using the quadratic formula: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-(-1)\pm \sqrt{(-1)^2-4\times 1\times 8}}{2\times 1} = \dfrac{1\pm \sqrt{1-32}}{2} = \dfrac{1\pm \sqrt{-31}}{2} = \dfrac{1\pm \sqrt{31}i}{2}$
Therefore the solutions are $x_{1}=\dfrac{1 + \sqrt{31}i}{2}$ and $x_{2}=\dfrac{1- \sqrt{31}i}{2}$
