(q+5)(q-4)=0

Solution for (q+5)(q-4)=0 equation:

(q+5)(q-4)=0

We multiply parentheses ..

(+q^2-4q+5q-20)=0

We get rid of parentheses

q^2-4q+5q-20=0

We add all the numbers together, and all the variables

q^2+q-20=0

a = 1; b = 1; c = -20;
Δ = b2-4ac
Δ = 12-4·1·(-20)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*1}=\frac{-10}{2}
=-5
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*1}=\frac{8}{2}
=4
$