(3q+20)(-q+4)=0

Solution for (3q+20)(-q+4)=0 equation:

(3q+20)(-q+4)=0

We add all the numbers together, and all the variables

(3q+20)(-1q+4)=0

We multiply parentheses ..

(-3q^2+12q-20q+80)=0

We get rid of parentheses

-3q^2+12q-20q+80=0

We add all the numbers together, and all the variables

-3q^2-8q+80=0

a = -3; b = -8; c = +80;
Δ = b2-4ac
Δ = -82-4·(-3)·80
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-32}{2*-3}=\frac{-24}{-6}
=+4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+32}{2*-3}=\frac{40}{-6}
=-6+2/3
$