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

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

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

We multiply parentheses ..

(+4q^2+5q+20q+25)=0

We get rid of parentheses

4q^2+5q+20q+25=0

We add all the numbers together, and all the variables

4q^2+25q+25=0

a = 4; b = 25; c = +25;
Δ = b2-4ac
Δ = 252-4·4·25
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-15}{2*4}=\frac{-40}{8}
=-5
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+15}{2*4}=\frac{-10}{8}
=-1+1/4
$