3q^2=-36+39q

Solution for 3q^2=-36+39q equation:

3q^2=-36+39q

We move all terms to the left:

3q^2-(-36+39q)=0

We add all the numbers together, and all the variables

3q^2-(39q-36)=0

We get rid of parentheses

3q^2-39q+36=0

a = 3; b = -39; c = +36;
Δ = b2-4ac
Δ = -392-4·3·36
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-33}{2*3}=\frac{6}{6}
=1
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+33}{2*3}=\frac{72}{6}
=12
$