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

Solution for (8-z)(4z-5)=0 equation:

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

We add all the numbers together, and all the variables

(-1z+8)(4z-5)=0

We multiply parentheses ..

(-4z^2+5z+32z-40)=0

We get rid of parentheses

-4z^2+5z+32z-40=0

We add all the numbers together, and all the variables

-4z^2+37z-40=0

a = -4; b = 37; c = -40;
Δ = b2-4ac
Δ = 372-4·(-4)·(-40)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-27}{2*-4}=\frac{-64}{-8}
=+8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+27}{2*-4}=\frac{-10}{-8}
=1+1/4
$