R=(20+x)*(300-10x)

Solution for R=(20+x)*(300-10x) equation:

=(20+R)(300-10R)

We move all terms to the left:

-((20+R)(300-10R))=0

We add all the numbers together, and all the variables

-((R+20)(-10R+300))=0

We multiply parentheses ..

-((-10R^2+300R-200R+6000))=0


We calculate terms in parentheses: -((-10R^2+300R-200R+6000)), so:

(-10R^2+300R-200R+6000)

We get rid of parentheses

-10R^2+300R-200R+6000

We add all the numbers together, and all the variables

-10R^2+100R+6000

Back to the equation:

-(-10R^2+100R+6000)


We get rid of parentheses

10R^2-100R-6000=0

a = 10; b = -100; c = -6000;
Δ = b2-4ac
Δ = -1002-4·10·(-6000)
Δ = 250000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{250000}=500$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-500}{2*10}=\frac{-400}{20}
=-20
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+500}{2*10}=\frac{600}{20}
=30
$