R=(3+1x)(800-100x)

Solution for R=(3+1x)(800-100x) equation:

=(3+1R)(800-100R)

We move all terms to the left:

-((3+1R)(800-100R))=0

We add all the numbers together, and all the variables

-((R+3)(-100R+800))=0

We multiply parentheses ..

-((-100R^2+800R-300R+2400))=0


We calculate terms in parentheses: -((-100R^2+800R-300R+2400)), so:

(-100R^2+800R-300R+2400)

We get rid of parentheses

-100R^2+800R-300R+2400

We add all the numbers together, and all the variables

-100R^2+500R+2400

Back to the equation:

-(-100R^2+500R+2400)


We get rid of parentheses

100R^2-500R-2400=0

a = 100; b = -500; c = -2400;
Δ = b2-4ac
Δ = -5002-4·100·(-2400)
Δ = 1210000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1210000}=1100$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-500)-1100}{2*100}=\frac{-600}{200}
=-3
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-500)+1100}{2*100}=\frac{1600}{200}
=8
$