Answer
{$\dfrac{4}{3}-\dfrac{\sqrt{13}}{3},\dfrac{4}{3} + \dfrac{\sqrt{13}}{3}$} or, {$\dfrac{4-\sqrt{13}}{3},\dfrac{4+\sqrt{13}}{3}$}
Work Step by Step
Given: $3x^2-8x+1=0$
This can be re-written as: $x^2-\dfrac{8}{3}x=\dfrac{-1}{3}$
We will have to add both sides $(-\dfrac{4}{3})^2$ to complete the square.
Thus, $x^2-\dfrac{8}{3}x+(-\dfrac{4}{3})^2=\dfrac{-1}{3}+(-\dfrac{4}{3})^2$
or, $(x-\dfrac{4}{3})^2=\dfrac{13}{9}$
or, $(x-\dfrac{4}{3})=\pm \dfrac{\sqrt{13}}{3}$
or, $x=\dfrac{4}{3} \pm \dfrac{\sqrt{13}}{3}$
Hence, our desired solution set is {$\dfrac{4}{3}-\dfrac{\sqrt{13}}{3},\dfrac{4}{3} + \dfrac{\sqrt{13}}{3}$} or, {$\dfrac{4-\sqrt{13}}{3},\dfrac{4+\sqrt{13}}{3}$}
