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

Solution for (6-z)(3z+1)=0 equation:

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

We add all the numbers together, and all the variables

(-1z+6)(3z+1)=0

We multiply parentheses ..

(-3z^2-1z+18z+6)=0

We get rid of parentheses

-3z^2-1z+18z+6=0

We add all the numbers together, and all the variables

-3z^2+17z+6=0

a = -3; b = 17; c = +6;
Δ = b2-4ac
Δ = 172-4·(-3)·6
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-19}{2*-3}=\frac{-36}{-6}
=+6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+19}{2*-3}=\frac{2}{-6}
=-1/3
$