D=(940+25x)(218-5x)

Solution for D=(940+25x)(218-5x) equation:

=(940+25D)(218-5D)

We move all terms to the left:

-((940+25D)(218-5D))=0

We add all the numbers together, and all the variables

-((25D+940)(-5D+218))=0

We multiply parentheses ..

-((-125D^2+5450D-4700D+204920))=0


We calculate terms in parentheses: -((-125D^2+5450D-4700D+204920)), so:

(-125D^2+5450D-4700D+204920)

We get rid of parentheses

-125D^2+5450D-4700D+204920

We add all the numbers together, and all the variables

-125D^2+750D+204920

Back to the equation:

-(-125D^2+750D+204920)


We get rid of parentheses

125D^2-750D-204920=0

a = 125; b = -750; c = -204920;
Δ = b2-4ac
Δ = -7502-4·125·(-204920)
Δ = 103022500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{103022500}=10150$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-750)-10150}{2*125}=\frac{-9400}{250}
=-37+3/5
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-750)+10150}{2*125}=\frac{10900}{250}
=43+3/5
$