## Precalculus (6th Edition) Blitzer

The equation has the solutions $\frac{5+\sqrt{13}}{6},\frac{5-\sqrt{13}}{6}$.
The provided quadratic equation is $3{{x}^{2}}-5x+1=0$. Here, $a=3$ , $b=-5$ , and $c=1$ \begin{align} & D={{\left( -5 \right)}^{2}}-4\left( 3 \right)\left( 1 \right) \\ & =25-12 \\ & =13 \end{align} Therefore, \begin{align} & x=\frac{-\left( -5 \right)\pm \sqrt{13}}{2\left( 3 \right)} \\ & =\frac{5\pm \sqrt{13}}{6} \end{align} Thus, $x=\frac{5+\sqrt{13}}{6}$ or $x=\frac{5-\sqrt{13}}{6}$.