(4q+4)(3q-8)=0

Solution for (4q+4)(3q-8)=0 equation:

(4q+4)(3q-8)=0

We multiply parentheses ..

(+12q^2-32q+12q-32)=0

We get rid of parentheses

12q^2-32q+12q-32=0

We add all the numbers together, and all the variables

12q^2-20q-32=0

a = 12; b = -20; c = -32;
Δ = b2-4ac
Δ = -202-4·12·(-32)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-44}{2*12}=\frac{-24}{24}
=-1
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+44}{2*12}=\frac{64}{24}
=2+2/3
$