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

Solution for (2-z)(3z-1)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3z^2+7z-2=0

a = -3; b = 7; c = -2;
Δ = b2-4ac
Δ = 72-4·(-3)·(-2)
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{-(7)-5}{2*-3}=\frac{-12}{-6}
=+2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*-3}=\frac{-2}{-6}
=1/3
$