5+3(y-4)=5(y+2)y

Solution for 5+3(y-4)=5(y+2)y equation:

5+3(y-4)=5(y+2)y

We move all terms to the left:

5+3(y-4)-(5(y+2)y)=0

We multiply parentheses

3y-(5(y+2)y)-12+5=0


We calculate terms in parentheses: -(5(y+2)y), so:

5(y+2)y

We multiply parentheses

5y^2+10y

Back to the equation:

-(5y^2+10y)


We add all the numbers together, and all the variables

3y-(5y^2+10y)-7=0

We get rid of parentheses

-5y^2+3y-10y-7=0

We add all the numbers together, and all the variables

-5y^2-7y-7=0

a = -5; b = -7; c = -7;
Δ = b2-4ac
Δ = -72-4·(-5)·(-7)
Δ = -91
Delta is less than zero, so there is no solution for the equation