(2-z)(3z-5)=0

Solution for (2-z)(3z-5)=0 equation:

(2-z)(3z-5)=0

We add all the numbers together, and all the variables

(-1z+2)(3z-5)=0

We multiply parentheses ..

(-3z^2+5z+6z-10)=0

We get rid of parentheses

-3z^2+5z+6z-10=0

We add all the numbers together, and all the variables

-3z^2+11z-10=0

a = -3; b = 11; c = -10;
Δ = b2-4ac
Δ = 112-4·(-3)·(-10)
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{-(11)-1}{2*-3}=\frac{-12}{-6}
=+2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*-3}=\frac{-10}{-6}
=1+2/3
$