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

Solution for (5-z)(4z-9)=0 equation:

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

We add all the numbers together, and all the variables

(-1z+5)(4z-9)=0

We multiply parentheses ..

(-4z^2+9z+20z-45)=0

We get rid of parentheses

-4z^2+9z+20z-45=0

We add all the numbers together, and all the variables

-4z^2+29z-45=0

a = -4; b = 29; c = -45;
Δ = b2-4ac
Δ = 292-4·(-4)·(-45)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-11}{2*-4}=\frac{-40}{-8}
=+5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+11}{2*-4}=\frac{-18}{-8}
=2+1/4
$