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

Solution for (5-z)(2z=9)=0 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

(-1z+5)(2z-9)=0

We multiply parentheses ..

(-2z^2+9z+10z-45)=0

We get rid of parentheses

-2z^2+9z+10z-45=0

We add all the numbers together, and all the variables

-2z^2+19z-45=0

a = -2; b = 19; c = -45;
Δ = b2-4ac
Δ = 192-4·(-2)·(-45)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*-2}=\frac{-20}{-4}
=+5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*-2}=\frac{-18}{-4}
=4+1/2
$