-.03q^2+1.2q=6

Solution for -.03q^2+1.2q=6 equation:

-.03q^2+1.2q=6

We move all terms to the left:

-.03q^2+1.2q-(6)=0

We add all the numbers together, and all the variables

-0.03q^2+1.2q-6=0

a = -0.03; b = 1.2; c = -6;
Δ = b2-4ac
Δ = 1.22-4·(-0.03)·(-6)
Δ = 0.72
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{-(1.2)-\sqrt{0.72}}{2*-0.03}=\frac{-1.2-\sqrt{0.72}}{-0.06}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{0.72}}{2*-0.03}=\frac{-1.2+\sqrt{0.72}}{-0.06}
$