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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4z^2+19z+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:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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