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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-5z^2+20z+9z-36)=0

We get rid of parentheses

-5z^2+20z+9z-36=0

We add all the numbers together, and all the variables

-5z^2+29z-36=0

a = -5; b = 29; c = -36;
Δ = b2-4ac
Δ = 292-4·(-5)·(-36)
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*-5}=\frac{-40}{-10}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+11}{2*-5}=\frac{-18}{-10}
=1+4/5
$