2q(q+1)=0

Solution for 2q(q+1)=0 equation:

2q(q+1)=0

We multiply parentheses

2q^2+2q=0

a = 2; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·2·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*2}=\frac{-4}{4}
=-1
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*2}=\frac{0}{4}
=0
$