6(2y+6)=4(9+3y)12y+36=36+12y12y=12y0=0

Solution for 6(2y+6)=4(9+3y)12y+36=36+12y12y=12y0=0 equation:

6(2y+6)=4(9+3y)12y+36=36+12y^12y=12y^0=0

We move all terms to the left:

6(2y+6)-(4(9+3y)12y+36)=0

We add all the numbers together, and all the variables

6(2y+6)-(4(3y+9)12y+36)=0

We multiply parentheses

12y-(4(3y+9)12y+36)+36=0


We calculate terms in parentheses: -(4(3y+9)12y+36), so:

4(3y+9)12y+36

We multiply parentheses

144y^2+432y+36

Back to the equation:

-(144y^2+432y+36)


We get rid of parentheses

-144y^2+12y-432y-36+36=0

We add all the numbers together, and all the variables

-144y^2-420y=0

a = -144; b = -420; c = 0;
Δ = b2-4ac
Δ = -4202-4·(-144)·0
Δ = 176400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{176400}=420$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-420)-420}{2*-144}=\frac{0}{-288}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-420)+420}{2*-144}=\frac{840}{-288}
=-2+11/12
$