(4z+4)(3z-2)=0

Solution for (4z+4)(3z-2)=0 equation:

(4z+4)(3z-2)=0

We multiply parentheses ..

(+12z^2-8z+12z-8)=0

We get rid of parentheses

12z^2-8z+12z-8=0

We add all the numbers together, and all the variables

12z^2+4z-8=0

a = 12; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·12·(-8)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*12}=\frac{-24}{24}
=-1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*12}=\frac{16}{24}
=2/3
$