R=(1,600+50x)(50-x)

Solution for R=(1,600+50x)(50-x) equation:

=(1.600+50R)(50-R)

We move all terms to the left:

-((1.600+50R)(50-R))=0

We add all the numbers together, and all the variables

-((50R+1.6)(-1R+50))=0

We multiply parentheses ..

-((-50R^2+2500R-1.6R+80))=0


We calculate terms in parentheses: -((-50R^2+2500R-1.6R+80)), so:

(-50R^2+2500R-1.6R+80)

We get rid of parentheses

-50R^2+2500R-1.6R+80

We add all the numbers together, and all the variables

-50R^2+2498.4R+80

Back to the equation:

-(-50R^2+2498.4R+80)


We get rid of parentheses

50R^2-2498.4R-80=0

a = 50; b = -2498.4; c = -80;
Δ = b2-4ac
Δ = -2498.42-4·50·(-80)
Δ = 6258002.56
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{-(-2498.4)-\sqrt{6258002.56}}{2*50}=\frac{2498.4-\sqrt{6258002.56}}{100}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2498.4)+\sqrt{6258002.56}}{2*50}=\frac{2498.4+\sqrt{6258002.56}}{100}
$