-3q2+118.5q-328.5=0

Solution for -3q2+118.5q-328.5=0 equation:

-3q^2+118.5q-328.5=0

a = -3; b = 118.5; c = -328.5;
Δ = b2-4ac
Δ = 118.52-4·(-3)·(-328.5)
Δ = 10100.25
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{-(118.5)-\sqrt{10100.25}}{2*-3}=\frac{-118.5-\sqrt{10100.25}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(118.5)+\sqrt{10100.25}}{2*-3}=\frac{-118.5+\sqrt{10100.25}}{-6}
$