-0.01q^2+0.9q+7=0

Solution for -0.01q^2+0.9q+7=0 equation:

-0.01q^2+0.9q+7=0

a = -0.01; b = 0.9; c = +7;
Δ = b2-4ac
Δ = 0.92-4·(-0.01)·7
Δ = 1.09
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{-(0.9)-\sqrt{1.09}}{2*-0.01}=\frac{-0.9-\sqrt{1.09}}{-0.02}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.9)+\sqrt{1.09}}{2*-0.01}=\frac{-0.9+\sqrt{1.09}}{-0.02}
$