(8x+1)(5x-1)=0

Solution for (8x+1)(5x-1)=0 equation:

(8x+1)(5x-1)=0

We multiply parentheses ..

(+40x^2-8x+5x-1)=0

We get rid of parentheses

40x^2-8x+5x-1=0

We add all the numbers together, and all the variables

40x^2-3x-1=0

a = 40; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·40·(-1)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*40}=\frac{-10}{80}
=-1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*40}=\frac{16}{80}
=1/5
$