0=-120q+612+3q^2

Solution for 0=-120q+612+3q^2 equation:

0=-120q+612+3q^2

We move all terms to the left:

0-(-120q+612+3q^2)=0

We add all the numbers together, and all the variables

-(-120q+612+3q^2)=0

We get rid of parentheses

-3q^2+120q-612=0

a = -3; b = 120; c = -612;
Δ = b2-4ac
Δ = 1202-4·(-3)·(-612)
Δ = 7056
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7056}=84$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-84}{2*-3}=\frac{-204}{-6}
=+34
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+84}{2*-3}=\frac{-36}{-6}
=+6
$