we have the equation
3y^2 - 6y = 1
equate to zero
3y^2 - 6y - 1=0
a=3
b=-6
c=-1
substitute the given values in the formula
[tex]y=\frac{-(-6)\pm\sqrt[]{-6^2-4(3)(-1)}}{2(3)}[/tex][tex]y=\frac{6\pm\sqrt[]{48}}{6}[/tex]simplify
[tex]y=\frac{6\pm4\sqrt[]{3}}{6}[/tex]the solutions for y are
[tex]y=1+\frac{2\sqrt[]{3}}{3}[/tex]and
[tex]y=1-\frac{2\sqrt[]{3}}{3}[/tex]