2(3q-7)(2q+2)=0

Solution for 2(3q-7)(2q+2)=0 equation:

2(3q-7)(2q+2)=0

We multiply parentheses ..

2(+6q^2+6q-14q-14)=0

We multiply parentheses

12q^2+12q-28q-28=0

We add all the numbers together, and all the variables

12q^2-16q-28=0

a = 12; b = -16; c = -28;
Δ = b2-4ac
Δ = -162-4·12·(-28)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-40}{2*12}=\frac{-24}{24}
=-1
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+40}{2*12}=\frac{56}{24}
=2+1/3
$