Answer
the solution set is\[\left\{ \frac{5\pm \sqrt{13}}{6} \right\}\].
Work Step by Step
The given equation can be written as\[3{{x}^{2}}-5x+1=0\].
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=3,\,\,b=-5,\text{ and }c=1\]
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{5\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 3\times 1}}{2\times 3} \\
& =\frac{5\pm \sqrt{25-12}}{6} \\
& =\frac{5\pm \sqrt{13}}{6}
\end{align}\]
Hence, the solution set is\[\left\{ \frac{5\pm \sqrt{13}}{6} \right\}\].
