3y=2(2+2y)3y=4+4y-y=4y=

Solution for 3y=2(2+2y)3y=4+4y-y=4y= equation:

3y=2(2+2y)3y=4+4y-y=4y=

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(2(2y+2)3y), so:

2(2y+2)3y

We multiply parentheses

12y^2+12y

Back to the equation:

-(12y^2+12y)


We get rid of parentheses

-12y^2+3y-12y=0

We add all the numbers together, and all the variables

-12y^2-9y=0

a = -12; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-12)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-12}=\frac{0}{-24}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-12}=\frac{18}{-24}
=-3/4
$