R(x)=(120+5x)(300-20x)

Solution for R(x)=(120+5x)(300-20x) equation:

(R)=(120+5R)(300-20R)

We move all terms to the left:

(R)-((120+5R)(300-20R))=0

We add all the numbers together, and all the variables

R-((5R+120)(-20R+300))=0

We multiply parentheses ..

-((-100R^2+1500R-2400R+36000))+R=0


We calculate terms in parentheses: -((-100R^2+1500R-2400R+36000)), so:

(-100R^2+1500R-2400R+36000)

We get rid of parentheses

-100R^2+1500R-2400R+36000

We add all the numbers together, and all the variables

-100R^2-900R+36000

Back to the equation:

-(-100R^2-900R+36000)


We get rid of parentheses

100R^2+900R+R-36000=0

We add all the numbers together, and all the variables

100R^2+901R-36000=0

a = 100; b = 901; c = -36000;
Δ = b2-4ac
Δ = 9012-4·100·(-36000)
Δ = 15211801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(901)-\sqrt{15211801}}{2*100}=\frac{-901-\sqrt{15211801}}{200}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(901)+\sqrt{15211801}}{2*100}=\frac{-901+\sqrt{15211801}}{200}
$