(3y-1)(y+1)=3y

Solution for (3y-1)(y+1)=3y equation:

(3y-1)(y+1)=3y

We move all terms to the left:

(3y-1)(y+1)-(3y)=0

We add all the numbers together, and all the variables

-3y+(3y-1)(y+1)=0

We multiply parentheses ..

(+3y^2+3y-1y-1)-3y=0

We get rid of parentheses

3y^2+3y-1y-3y-1=0

We add all the numbers together, and all the variables

3y^2-1y-1=0

a = 3; b = -1; c = -1;
Δ = b2-4ac
Δ = -12-4·3·(-1)
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{13}}{2*3}=\frac{1-\sqrt{13}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{13}}{2*3}=\frac{1+\sqrt{13}}{6}
$