(Y+4)^2=(4-y)(4+y)

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

(+4)^2=(4-Y)(4+Y)

We move all terms to the left:

(+4)^2-((4-Y)(4+Y))=0

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

-((-1Y+4)(Y+4))+16=0

We multiply parentheses ..

-((-1Y^2-4Y+4Y+16))+16=0


We calculate terms in parentheses: -((-1Y^2-4Y+4Y+16)), so:

(-1Y^2-4Y+4Y+16)

We get rid of parentheses

-1Y^2-4Y+4Y+16

We add all the numbers together, and all the variables

-1Y^2+16

Back to the equation:

-(-1Y^2+16)


We get rid of parentheses

1Y^2-16+16=0

We add all the numbers together, and all the variables

Y^2=0

a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$Y=\frac{-b}{2a}=\frac{0}{2}=0$