R=(50-2x)(40-2x)

Solution for R=(50-2x)(40-2x) equation:

=(50-2R)(40-2R)

We move all terms to the left:

-((50-2R)(40-2R))=0

We add all the numbers together, and all the variables

-((-2R+50)(-2R+40))=0

We multiply parentheses ..

-((+4R^2-80R-100R+2000))=0


We calculate terms in parentheses: -((+4R^2-80R-100R+2000)), so:

(+4R^2-80R-100R+2000)

We get rid of parentheses

4R^2-80R-100R+2000

We add all the numbers together, and all the variables

4R^2-180R+2000

Back to the equation:

-(4R^2-180R+2000)


We get rid of parentheses

-4R^2+180R-2000=0

a = -4; b = 180; c = -2000;
Δ = b2-4ac
Δ = 1802-4·(-4)·(-2000)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(180)-20}{2*-4}=\frac{-200}{-8}
=+25
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(180)+20}{2*-4}=\frac{-160}{-8}
=+20
$