(8c-5)(4c-1)=0

Solution for (8c-5)(4c-1)=0 equation:

(8c-5)(4c-1)=0

We multiply parentheses ..

(+32c^2-8c-20c+5)=0

We get rid of parentheses

32c^2-8c-20c+5=0

We add all the numbers together, and all the variables

32c^2-28c+5=0

a = 32; b = -28; c = +5;
Δ = b2-4ac
Δ = -282-4·32·5
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-12}{2*32}=\frac{16}{64}
=1/4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+12}{2*32}=\frac{40}{64}
=5/8
$