-0.03q^2+3q-48=0

Solution for -0.03q^2+3q-48=0 equation:

-0.03q^2+3q-48=0

a = -0.03; b = 3; c = -48;
Δ = b2-4ac
Δ = 32-4·(-0.03)·(-48)
Δ = 3.24
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{-(3)-\sqrt{3.24}}{2*-0.03}=\frac{-3-\sqrt{3.24}}{-0.06}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{3.24}}{2*-0.03}=\frac{-3+\sqrt{3.24}}{-0.06}
$