Answer
{$\dfrac{1- \sqrt{7}}{6},\dfrac{1 + \sqrt{7}}{6}$}
Work Step by Step
Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
Since, $6x^2-2x-1=0$
Thus, $x=\dfrac{-(-2) \pm \sqrt{(-2)^2-4(6)(-1)}}{2(6)}$
or, $x=\dfrac{2 \pm \sqrt{28}}{12}$
or, $x=\dfrac{1 \pm \sqrt{7}}{6}$
or, $x=\dfrac{1- \sqrt{7}}{6},\dfrac{1 + \sqrt{7}}{6}$
Our solution set is: {$\dfrac{1- \sqrt{7}}{6},\dfrac{1 + \sqrt{7}}{6}$}
