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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+2x^2+5x-16x-40))=0


We calculate terms in parentheses: -((+2x^2+5x-16x-40)), so:

(+2x^2+5x-16x-40)

We get rid of parentheses

2x^2+5x-16x-40

We add all the numbers together, and all the variables

2x^2-11x-40

Back to the equation:

-(2x^2-11x-40)


We get rid of parentheses

-2x^2+11x+40=0

a = -2; b = 11; c = +40;
Δ = b2-4ac
Δ = 112-4·(-2)·40
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*-2}=\frac{-32}{-4}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*-2}=\frac{10}{-4}
=-2+1/2
$