0.002q^2+0.5q=0

Solution for 0.002q^2+0.5q=0 equation:

0.002q^2+0.5q=0

a = 0.002; b = 0.5; c = 0;
Δ = b2-4ac
Δ = 0.52-4·0.002·0
Δ = 0.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{0.25}}{2*0.002}=\frac{-0.5-\sqrt{0.25}}{0.004}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.25}}{2*0.002}=\frac{-0.5+\sqrt{0.25}}{0.004}
$