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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3z^2+z+2=0

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