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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((-2z^2+4z+9z-18))=0


We calculate terms in parentheses: -((-2z^2+4z+9z-18)), so:

(-2z^2+4z+9z-18)

We get rid of parentheses

-2z^2+4z+9z-18

We add all the numbers together, and all the variables

-2z^2+13z-18

Back to the equation:

-(-2z^2+13z-18)


We get rid of parentheses

2z^2-13z+18=0

a = 2; b = -13; c = +18;
Δ = b2-4ac
Δ = -132-4·2·18
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*2}=\frac{8}{4}
=2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*2}=\frac{18}{4}
=4+1/2
$