(4+y)(4+y)=16-y^2

Solution for (4+y)(4+y)=16-y^2 equation:

(4+y)(4+y)=16-y^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

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

We get rid of parentheses

y^2+y^2+4y+4y+16-16=0

We add all the numbers together, and all the variables

2y^2+8y=0

a = 2; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·2·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*2}=\frac{-16}{4}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*2}=\frac{0}{4}
=0
$