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

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

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

We multiply parentheses ..

(+20x^2+40x-16x-32)=0

We get rid of parentheses

20x^2+40x-16x-32=0

We add all the numbers together, and all the variables

20x^2+24x-32=0

a = 20; b = 24; c = -32;
Δ = b2-4ac
Δ = 242-4·20·(-32)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-56}{2*20}=\frac{-80}{40}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+56}{2*20}=\frac{32}{40}
=4/5
$