## College Algebra (6th Edition)

$\{\frac{7+\sqrt 37}{6},\frac{7-\sqrt 37}{6}\}$
$3x^{2}-7x+1=0$ 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 $x={\frac{7-\sqrt 37}{6}}$