3(q^2-3q)=0

Solution for 3(q^2-3q)=0 equation:

3(q^2-3q)=0

We multiply parentheses

3q^2-9q=0

a = 3; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·3·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*3}=\frac{0}{6}
=0
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*3}=\frac{18}{6}
=3
$