(3q+4)(4q+2)=0

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

(3q+4)(4q+2)=0

We multiply parentheses ..

(+12q^2+6q+16q+8)=0

We get rid of parentheses

12q^2+6q+16q+8=0

We add all the numbers together, and all the variables

12q^2+22q+8=0

a = 12; b = 22; c = +8;
Δ = b2-4ac
Δ = 222-4·12·8
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-10}{2*12}=\frac{-32}{24}
=-1+1/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+10}{2*12}=\frac{-12}{24}
=-1/2
$