(6q+3)q=0

Solution for (6q+3)q=0 equation:

(6q+3)q=0

We multiply parentheses

6q^2+3q=0

a = 6; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·6·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*6}=\frac{-6}{12}
=-1/2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*6}=\frac{0}{12}
=0
$