-3(y+4)4y=-3y-12+4y

Solution for -3(y+4)4y=-3y-12+4y equation:

-3(y+4)4y=-3y-12+4y

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-12y^2-48y-(y-12)=0

We get rid of parentheses

-12y^2-48y-y+12=0

We add all the numbers together, and all the variables

-12y^2-49y+12=0

a = -12; b = -49; c = +12;
Δ = b2-4ac
Δ = -492-4·(-12)·12
Δ = 2977
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{-(-49)-\sqrt{2977}}{2*-12}=\frac{49-\sqrt{2977}}{-24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{2977}}{2*-12}=\frac{49+\sqrt{2977}}{-24}
$