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

Solution for (5-y)(4y+1)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4y^2-1y+20y+5)=0

We get rid of parentheses

-4y^2-1y+20y+5=0

We add all the numbers together, and all the variables

-4y^2+19y+5=0

a = -4; b = 19; c = +5;
Δ = b2-4ac
Δ = 192-4·(-4)·5
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*-4}=\frac{-40}{-8}
=+5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*-4}=\frac{2}{-8}
=-1/4
$