(4-z)(5z-2)=0

Solution for (4-z)(5z-2)=0 equation:

(4-z)(5z-2)=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-5z^2+2z+20z-8)=0

We get rid of parentheses

-5z^2+2z+20z-8=0

We add all the numbers together, and all the variables

-5z^2+22z-8=0

a = -5; b = 22; c = -8;
Δ = b2-4ac
Δ = 222-4·(-5)·(-8)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-18}{2*-5}=\frac{-40}{-10}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+18}{2*-5}=\frac{-4}{-10}
=2/5
$