(4+y)(4y)=0

Solution for (4+y)(4y)=0 equation:

(4+y)(4y)=0

We add all the numbers together, and all the variables

(y+4)4y=0

We multiply parentheses

4y^2+16y=0

a = 4; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·4·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*4}=\frac{-32}{8}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*4}=\frac{0}{8}
=0
$