(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+z+18z-6)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3z^2+19z-6=0

a = -3; b = 19; c = -6;
Δ = b2-4ac
Δ = 192-4·(-3)·(-6)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*-3}=\frac{-36}{-6}
=+6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*-3}=\frac{-2}{-6}
=1/3
$