3q(q-6)=-24

Solution for 3q(q-6)=-24 equation:

3q(q-6)=-24

We move all terms to the left:

3q(q-6)-(-24)=0

We add all the numbers together, and all the variables

3q(q-6)+24=0

We multiply parentheses

3q^2-18q+24=0

a = 3; b = -18; c = +24;
Δ = b2-4ac
Δ = -182-4·3·24
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*3}=\frac{12}{6}
=2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*3}=\frac{24}{6}
=4
$