(z-12)(4z+6)=0

Solution for (z-12)(4z+6)=0 equation:

(z-12)(4z+6)=0

We multiply parentheses ..

(+4z^2+6z-48z-72)=0

We get rid of parentheses

4z^2+6z-48z-72=0

We add all the numbers together, and all the variables

4z^2-42z-72=0

a = 4; b = -42; c = -72;
Δ = b2-4ac
Δ = -422-4·4·(-72)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-54}{2*4}=\frac{-12}{8}
=-1+1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+54}{2*4}=\frac{96}{8}
=12
$