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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-12y^2-1y+16y-(-2(y-4))=0


We calculate terms in parentheses: -(-2(y-4)), so:

-2(y-4)

We multiply parentheses

-2y+8

Back to the equation:

-(-2y+8)


We add all the numbers together, and all the variables

-12y^2+15y-(-2y+8)=0

We get rid of parentheses

-12y^2+15y+2y-8=0

We add all the numbers together, and all the variables

-12y^2+17y-8=0

a = -12; b = 17; c = -8;
Δ = b2-4ac
Δ = 172-4·(-12)·(-8)
Δ = -95
Delta is less than zero, so there is no solution for the equation