(8-z)(4z-9)=0

Solution for (8-z)(4z-9)=0 equation:

(8-z)(4z-9)=0

We add all the numbers together, and all the variables

(-1z+8)(4z-9)=0

We multiply parentheses ..

(-4z^2+9z+32z-72)=0

We get rid of parentheses

-4z^2+9z+32z-72=0

We add all the numbers together, and all the variables

-4z^2+41z-72=0

a = -4; b = 41; c = -72;
Δ = b2-4ac
Δ = 412-4·(-4)·(-72)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-23}{2*-4}=\frac{-64}{-8}
=+8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+23}{2*-4}=\frac{-18}{-8}
=2+1/4
$