(3q+1)(q+7)=0

Solution for (3q+1)(q+7)=0 equation:

(3q+1)(q+7)=0

We multiply parentheses ..

(+3q^2+21q+q+7)=0

We get rid of parentheses

3q^2+21q+q+7=0

We add all the numbers together, and all the variables

3q^2+22q+7=0

a = 3; b = 22; c = +7;
Δ = b2-4ac
Δ = 222-4·3·7
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-20}{2*3}=\frac{-42}{6}
=-7
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+20}{2*3}=\frac{-2}{6}
=-1/3
$