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

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

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

We multiply parentheses ..

(+2x^2+10x-8x-40)=0

We get rid of parentheses

2x^2+10x-8x-40=0

We add all the numbers together, and all the variables

2x^2+2x-40=0

a = 2; b = 2; c = -40;
Δ = b2-4ac
Δ = 22-4·2·(-40)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-18}{2*2}=\frac{-20}{4}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+18}{2*2}=\frac{16}{4}
=4
$