Answer
The equation has the solutions $\frac{5+\sqrt{13}}{6},\frac{5-\sqrt{13}}{6}$.
Work Step by Step
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 $x=\frac{5-\sqrt{13}}{6}$.
