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

Solution for (4z-9)(1-z)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4z^2+13z-9=0

a = -4; b = 13; c = -9;
Δ = b2-4ac
Δ = 132-4·(-4)·(-9)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*-4}=\frac{-18}{-8}
=2+1/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*-4}=\frac{-8}{-8}
=1
$