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

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

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

We move all terms to the left:

-4(2-y)+3y-(3y(y-4))=0

We add all the numbers together, and all the variables

-4(-1y+2)+3y-(3y(y-4))=0

We add all the numbers together, and all the variables

3y-4(-1y+2)-(3y(y-4))=0

We multiply parentheses

3y+4y-(3y(y-4))-8=0


We calculate terms in parentheses: -(3y(y-4)), so:

3y(y-4)

We multiply parentheses

3y^2-12y

Back to the equation:

-(3y^2-12y)


We add all the numbers together, and all the variables

7y-(3y^2-12y)-8=0

We get rid of parentheses

-3y^2+7y+12y-8=0

We add all the numbers together, and all the variables

-3y^2+19y-8=0

a = -3; b = 19; c = -8;
Δ = b2-4ac
Δ = 192-4·(-3)·(-8)
Δ = 265
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{-(19)-\sqrt{265}}{2*-3}=\frac{-19-\sqrt{265}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{265}}{2*-3}=\frac{-19+\sqrt{265}}{-6}
$