3q(q-3)=12

Solution for 3q(q-3)=12 equation:

3q(q-3)=12

We move all terms to the left:

3q(q-3)-(12)=0

We multiply parentheses

3q^2-9q-12=0

a = 3; b = -9; c = -12;
Δ = b2-4ac
Δ = -92-4·3·(-12)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-15}{2*3}=\frac{-6}{6}
=-1
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+15}{2*3}=\frac{24}{6}
=4
$