Answer
$x_{1}=\dfrac{1 + \sqrt{35}i}{9}$ and $x_{2}=\dfrac{1- \sqrt{35}i}{9}$
Work Step by Step
Given $9a^2+4=2a \longrightarrow 9a^2-2a+4=0$
$a=9, \ b=-2, \ c=4$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-(-2) \pm \sqrt{(-2)^2-4\times 9\times 4}}{2\times 9} = \dfrac{2 \pm \sqrt{4-144}}{18} = \dfrac{2 \pm \sqrt{-140}}{18} = \dfrac{2 \pm \sqrt{140}i}{18} = \dfrac{2 \pm 2\sqrt{35}i}{18} = \dfrac{1 \pm \sqrt{35}i}{9}$
Therefore, the solutions are $x_{1}=\dfrac{1 + \sqrt{35}i}{9}$ and $x_{2}=\dfrac{1- \sqrt{35}i}{9}$
