(4x-8)(5x+2)=0

Solution for (4x-8)(5x+2)=0 equation:

(4x-8)(5x+2)=0

We multiply parentheses ..

(+20x^2+8x-40x-16)=0

We get rid of parentheses

20x^2+8x-40x-16=0

We add all the numbers together, and all the variables

20x^2-32x-16=0

a = 20; b = -32; c = -16;
Δ = b2-4ac
Δ = -322-4·20·(-16)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-48}{2*20}=\frac{-16}{40}
=-2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+48}{2*20}=\frac{80}{40}
=2
$