Answer
$\{\frac{7+\sqrt 37}{6},\frac{7-\sqrt 37}{6}\}$
Work Step by Step
$3x^{2}-7x+1=0$
Dividing both sides by $3$
$x^{2}-\frac{7}{3}x+\frac{1}{3}=0$
$x^{2}-\frac{7}{3}x=-\frac{1}{3}$
Adding the square of half the co-efficient of $x$ to both sides,
$x^{2}-\frac{7}{3}x+\frac{49}{36}=-\frac{1}{3}+\frac{49}{36}$
$(x-\frac{7}{6})^{2}=\frac{-12+49}{36}$
$(x-\frac{7}{6})^{2}=\frac{37}{36}$
$(x-\frac{7}{6})=±\sqrt \frac{37}{36}$
$x=\frac{7}{6}±\sqrt \frac{37}{36}$
$x=\frac{7}{6}±\frac{\sqrt 37}{6}$
$x={\frac{7+\sqrt 37}{6}}$ or $x={\frac{7-\sqrt 37}{6}}$
