3840=(100+5x)(40-2x)

Solution for 3840=(100+5x)(40-2x) equation:

3840=(100+5x)(40-2x)

We move all terms to the left:

3840-((100+5x)(40-2x))=0

We add all the numbers together, and all the variables

-((5x+100)(-2x+40))+3840=0

We multiply parentheses ..

-((-10x^2+200x-200x+4000))+3840=0


We calculate terms in parentheses: -((-10x^2+200x-200x+4000)), so:

(-10x^2+200x-200x+4000)

We get rid of parentheses

-10x^2+200x-200x+4000

We add all the numbers together, and all the variables

-10x^2+4000

Back to the equation:

-(-10x^2+4000)


We get rid of parentheses

10x^2-4000+3840=0

We add all the numbers together, and all the variables

10x^2-160=0

a = 10; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·10·(-160)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*10}=\frac{-80}{20}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*10}=\frac{80}{20}
=4
$