0=3q^2+10q-3000

Solution for 0=3q^2+10q-3000 equation:

0=3q^2+10q-3000

We move all terms to the left:

0-(3q^2+10q-3000)=0

We add all the numbers together, and all the variables

-(3q^2+10q-3000)=0

We get rid of parentheses

-3q^2-10q+3000=0

a = -3; b = -10; c = +3000;
Δ = b2-4ac
Δ = -102-4·(-3)·3000
Δ = 36100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36100}=190$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-190}{2*-3}=\frac{-180}{-6}
=+30
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+190}{2*-3}=\frac{200}{-6}
=-33+1/3
$