-0.2q^2+11q-30=0

Solution for -0.2q^2+11q-30=0 equation:

-0.2q^2+11q-30=0

a = -0.2; b = 11; c = -30;
Δ = b2-4ac
Δ = 112-4·(-0.2)·(-30)
Δ = 97
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{-(11)-\sqrt{97}}{2*-0.2}=\frac{-11-\sqrt{97}}{-0.4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{97}}{2*-0.2}=\frac{-11+\sqrt{97}}{-0.4}
$