5q(q+1)=0

Solution for 5q(q+1)=0 equation:

5q(q+1)=0

We multiply parentheses

5q^2+5q=0

a = 5; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·5·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*5}=\frac{-10}{10}
=-1
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*5}=\frac{0}{10}
=0
$