4y(3y-1)+4(3y-1)=0

Solution for 4y(3y-1)+4(3y-1)=0 equation:

4y(3y-1)+4(3y-1)=0

We multiply parentheses

12y^2-4y+12y-4=0

We add all the numbers together, and all the variables

12y^2+8y-4=0

a = 12; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·12·(-4)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*12}=\frac{-24}{24}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*12}=\frac{8}{24}
=1/3
$