-.3q+.002q^2=-.6q+.006q^2

Solution for -.3q+.002q^2=-.6q+.006q^2 equation:

-.3q+.002q^2=-.6q+.006q^2

We move all terms to the left:

-.3q+.002q^2-(-.6q+.006q^2)=0

We add all the numbers together, and all the variables

.002q^2-(-.6q+.006q^2)-0.3q=0

We get rid of parentheses

.002q^2-.006q^2+.6q-0.3q=0

We add all the numbers together, and all the variables

-0.004q^2+0.3q=0

a = -0.004; b = 0.3; c = 0;
Δ = b2-4ac
Δ = 0.32-4·(-0.004)·0
Δ = 0.09
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{-(0.3)-\sqrt{0.09}}{2*-0.004}=\frac{-0.3-\sqrt{0.09}}{-0.008}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{0.09}}{2*-0.004}=\frac{-0.3+\sqrt{0.09}}{-0.008}
$