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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4q^2-15q-25=0

a = 4; b = -15; c = -25;
Δ = b2-4ac
Δ = -152-4·4·(-25)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-25}{2*4}=\frac{-10}{8}
=-1+1/4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+25}{2*4}=\frac{40}{8}
=5
$