R=(800-100x)(12+4x)

Solution for R=(800-100x)(12+4x) equation:

=(800-100R)(12+4R)

We move all terms to the left:

-((800-100R)(12+4R))=0

We add all the numbers together, and all the variables

-((-100R+800)(4R+12))=0

We multiply parentheses ..

-((-400R^2-1200R+3200R+9600))=0


We calculate terms in parentheses: -((-400R^2-1200R+3200R+9600)), so:

(-400R^2-1200R+3200R+9600)

We get rid of parentheses

-400R^2-1200R+3200R+9600

We add all the numbers together, and all the variables

-400R^2+2000R+9600

Back to the equation:

-(-400R^2+2000R+9600)


We get rid of parentheses

400R^2-2000R-9600=0

a = 400; b = -2000; c = -9600;
Δ = b2-4ac
Δ = -20002-4·400·(-9600)
Δ = 19360000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19360000}=4400$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2000)-4400}{2*400}=\frac{-2400}{800}
=-3
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2000)+4400}{2*400}=\frac{6400}{800}
=8
$