-0.2q^2+30q+10=0

Solution for -0.2q^2+30q+10=0 equation:

-0.2q^2+30q+10=0

a = -0.2; b = 30; c = +10;
Δ = b2-4ac
Δ = 302-4·(-0.2)·10
Δ = 908
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{908}=\sqrt{4*227}=\sqrt{4}*\sqrt{227}=2\sqrt{227}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{227}}{2*-0.2}=\frac{-30-2\sqrt{227}}{-0.4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{227}}{2*-0.2}=\frac{-30+2\sqrt{227}}{-0.4}
$