(6-z)(3z-7)=0

Solution for (6-z)(3z-7)=0 equation:

(6-z)(3z-7)=0

We add all the numbers together, and all the variables

(-1z+6)(3z-7)=0

We multiply parentheses ..

(-3z^2+7z+18z-42)=0

We get rid of parentheses

-3z^2+7z+18z-42=0

We add all the numbers together, and all the variables

-3z^2+25z-42=0

a = -3; b = 25; c = -42;
Δ = b2-4ac
Δ = 252-4·(-3)·(-42)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-11}{2*-3}=\frac{-36}{-6}
=+6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+11}{2*-3}=\frac{-14}{-6}
=2+1/3
$