(3q+7)(4q+5)=0

Solution for (3q+7)(4q+5)=0 equation:

(3q+7)(4q+5)=0

We multiply parentheses ..

(+12q^2+15q+28q+35)=0

We get rid of parentheses

12q^2+15q+28q+35=0

We add all the numbers together, and all the variables

12q^2+43q+35=0

a = 12; b = 43; c = +35;
Δ = b2-4ac
Δ = 432-4·12·35
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-13}{2*12}=\frac{-56}{24}
=-2+1/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+13}{2*12}=\frac{-30}{24}
=-1+1/4
$