(q-5)(4q-1)=0

Solution for (q-5)(4q-1)=0 equation:

(q-5)(4q-1)=0

We multiply parentheses ..

(+4q^2-1q-20q+5)=0

We get rid of parentheses

4q^2-1q-20q+5=0

We add all the numbers together, and all the variables

4q^2-21q+5=0

a = 4; b = -21; c = +5;
Δ = b2-4ac
Δ = -212-4·4·5
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-19}{2*4}=\frac{2}{8}
=1/4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+19}{2*4}=\frac{40}{8}
=5
$