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

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

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

We multiply parentheses ..

(+3q^2+6q+7q+14)=0

We get rid of parentheses

3q^2+6q+7q+14=0

We add all the numbers together, and all the variables

3q^2+13q+14=0

a = 3; b = 13; c = +14;
Δ = b2-4ac
Δ = 132-4·3·14
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-1}{2*3}=\frac{-14}{6}
=-2+1/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+1}{2*3}=\frac{-12}{6}
=-2
$