R=(10+n)(250-5n)

Solution for R=(10+n)(250-5n) equation:

=(10+R)(250-5R)

We move all terms to the left:

-((10+R)(250-5R))=0

We add all the numbers together, and all the variables

-((R+10)(-5R+250))=0

We multiply parentheses ..

-((-5R^2+250R-50R+2500))=0


We calculate terms in parentheses: -((-5R^2+250R-50R+2500)), so:

(-5R^2+250R-50R+2500)

We get rid of parentheses

-5R^2+250R-50R+2500

We add all the numbers together, and all the variables

-5R^2+200R+2500

Back to the equation:

-(-5R^2+200R+2500)


We get rid of parentheses

5R^2-200R-2500=0

a = 5; b = -200; c = -2500;
Δ = b2-4ac
Δ = -2002-4·5·(-2500)
Δ = 90000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{90000}=300$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-200)-300}{2*5}=\frac{-100}{10}
=-10
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-200)+300}{2*5}=\frac{500}{10}
=50
$