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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

3z^2+3z-2z-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
$