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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2y-((-1y+4)(y+4))+8=0

We multiply parentheses ..

-((-1y^2-4y+4y+16))+2y+8=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+2y-16+8=0

We add all the numbers together, and all the variables

y^2+2y-8=0

a = 1; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·1·(-8)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*1}=\frac{-8}{2}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*1}=\frac{4}{2}
=2
$