Answer
Solution set = $\displaystyle \{\frac{3-\sqrt{6}}{3},\frac{3+\sqrt{6}}{3}\}.$
Work Step by Step
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Add $-6x+1$ to both sides, so we have $ax^{2}+bx+c=0.$
$ 3x^{2}-6x+1=0\quad\rightarrow \left\{\begin{array}{l}
a=3\\
b=-6\\
c=1
\end{array}\right.$
$x=\displaystyle \frac{6\pm\sqrt{(-6)^{2}-4(3)(1)}}{2(3)}=\frac{6\pm\sqrt{36-12}}{6}=\frac{6\pm\sqrt{24}}{6}$
$=\displaystyle \frac{6\pm\sqrt{4\cdot 6}}{6}=\frac{6\pm 2\sqrt{6}}{6}=\frac{2(3\pm\sqrt{6})}{6}$
$=\displaystyle \frac{3\pm\sqrt{6}}{3}$
Solution set = $\displaystyle \{\frac{3-\sqrt{6}}{3},\frac{3+\sqrt{6}}{3}\}.$
