0=3q^2+24q+20-164

Solution for 0=3q^2+24q+20-164 equation:

0=3q^2+24q+20-164

We move all terms to the left:

0-(3q^2+24q+20-164)=0

We add all the numbers together, and all the variables

-(3q^2+24q+20-164)=0

We get rid of parentheses

-3q^2-24q-20+164=0

We add all the numbers together, and all the variables

-3q^2-24q+144=0

a = -3; b = -24; c = +144;
Δ = b2-4ac
Δ = -242-4·(-3)·144
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-48}{2*-3}=\frac{-24}{-6}
=+4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+48}{2*-3}=\frac{72}{-6}
=-12
$