0=(3y^2)-6y+2

Solution for 0=(3y^2)-6y+2 equation:

0=(3y^2)-6y+2

We move all terms to the left:

0-((3y^2)-6y+2)=0

We add all the numbers together, and all the variables

-(3y^2-6y+2)=0

We get rid of parentheses

-3y^2+6y-2=0

a = -3; b = 6; c = -2;
Δ = b2-4ac
Δ = 62-4·(-3)·(-2)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{3}}{2*-3}=\frac{-6-2\sqrt{3}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{3}}{2*-3}=\frac{-6+2\sqrt{3}}{-6}
$