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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3z^2+9z-1z+3)=0

We get rid of parentheses

-3z^2+9z-1z+3=0

We add all the numbers together, and all the variables

-3z^2+8z+3=0

a = -3; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·(-3)·3
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-10}{2*-3}=\frac{-18}{-6}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+10}{2*-3}=\frac{2}{-6}
=-1/3
$