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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+4x^2-20x-8x+40)), so:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2-28x+40

Back to the equation:

-(4x^2-28x+40)


We get rid of parentheses

-4x^2+28x-40=0

a = -4; b = 28; c = -40;
Δ = b2-4ac
Δ = 282-4·(-4)·(-40)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-12}{2*-4}=\frac{-40}{-8}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+12}{2*-4}=\frac{-16}{-8}
=+2
$