-3q^2+20q+1000=0

Solution for -3q^2+20q+1000=0 equation:

-3q^2+20q+1000=0

a = -3; b = 20; c = +1000;
Δ = b2-4ac
Δ = 202-4·(-3)·1000
Δ = 12400
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:

$\sqrt{\Delta}=\sqrt{12400}=\sqrt{400*31}=\sqrt{400}*\sqrt{31}=20\sqrt{31}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{31}}{2*-3}=\frac{-20-20\sqrt{31}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{31}}{2*-3}=\frac{-20+20\sqrt{31}}{-6}
$