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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-5z^2+21z-4=0

a = -5; b = 21; c = -4;
Δ = b2-4ac
Δ = 212-4·(-5)·(-4)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*-5}=\frac{-40}{-10}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*-5}=\frac{-2}{-10}
=1/5
$