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

Solution for y.3(y+1)+4=2(y-1)=y+y equation:

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(2(y-1)), so:

2(y-1)

We multiply parentheses

2y-2

Back to the equation:

-(2y-2)


We get rid of parentheses

y^2+y-2y+2+4=0

We add all the numbers together, and all the variables

y^2-1y+6=0

a = 1; b = -1; c = +6;
Δ = b2-4ac
Δ = -12-4·1·6
Δ = -23
Delta is less than zero, so there is no solution for the equation