(3-z)(4z-6)=0

Solution for (3-z)(4z-6)=0 equation:

(3-z)(4z-6)=0

We add all the numbers together, and all the variables

(-1z+3)(4z-6)=0

We multiply parentheses ..

(-4z^2+6z+12z-18)=0

We get rid of parentheses

-4z^2+6z+12z-18=0

We add all the numbers together, and all the variables

-4z^2+18z-18=0

a = -4; b = 18; c = -18;
Δ = b2-4ac
Δ = 182-4·(-4)·(-18)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*-4}=\frac{-24}{-8}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*-4}=\frac{-12}{-8}
=1+1/2
$