31=0.02q^2+4q

Solution for 31=0.02q^2+4q equation:

31=0.02q^2+4q

We move all terms to the left:

31-(0.02q^2+4q)=0

We get rid of parentheses

-0.02q^2-4q+31=0

a = -0.02; b = -4; c = +31;
Δ = b2-4ac
Δ = -42-4·(-0.02)·31
Δ = 18.48
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{-(-4)-\sqrt{18.48}}{2*-0.02}=\frac{4-\sqrt{18.48}}{-0.04}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{18.48}}{2*-0.02}=\frac{4+\sqrt{18.48}}{-0.04}
$