(q+40)(q-20)=0

Solution for (q+40)(q-20)=0 equation:

(q+40)(q-20)=0

We multiply parentheses ..

(+q^2-20q+40q-800)=0

We get rid of parentheses

q^2-20q+40q-800=0

We add all the numbers together, and all the variables

q^2+20q-800=0

a = 1; b = 20; c = -800;
Δ = b2-4ac
Δ = 202-4·1·(-800)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-60}{2*1}=\frac{-80}{2}
=-40
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+60}{2*1}=\frac{40}{2}
=20
$