(4x-8)(5x+1)=0

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

(4x-8)(5x+1)=0

We multiply parentheses ..

(+20x^2+4x-40x-8)=0

We get rid of parentheses

20x^2+4x-40x-8=0

We add all the numbers together, and all the variables

20x^2-36x-8=0

a = 20; b = -36; c = -8;
Δ = b2-4ac
Δ = -362-4·20·(-8)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-44}{2*20}=\frac{-8}{40}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+44}{2*20}=\frac{80}{40}
=2
$