(2z-9)(3-z)=0

Solution for (2z-9)(3-z)=0 equation:

(2z-9)(3-z)=0

We add all the numbers together, and all the variables

(2z-9)(-1z+3)=0

We multiply parentheses ..

(-2z^2+6z+9z-27)=0

We get rid of parentheses

-2z^2+6z+9z-27=0

We add all the numbers together, and all the variables

-2z^2+15z-27=0

a = -2; b = 15; c = -27;
Δ = b2-4ac
Δ = 152-4·(-2)·(-27)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3}{2*-2}=\frac{-18}{-4}
=4+1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3}{2*-2}=\frac{-12}{-4}
=+3
$