(5y+Y)(4-y)=0

Solution for (5y+Y)(4-y)=0 equation:

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

We add all the numbers together, and all the variables

(+5y)(-1y+4)=0

We multiply parentheses ..

(-5y^2+20y)=0

We get rid of parentheses

-5y^2+20y=0

a = -5; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-5)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-5}=\frac{-40}{-10}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-5}=\frac{0}{-10}
=0
$