R(x)=(320+-20x)(70+5x)

Solution for R(x)=(320+-20x)(70+5x) equation:

(R)=(320+-20R)(70+5R)

We move all terms to the left:

(R)-((320+-20R)(70+5R))=0

We add all the numbers together, and all the variables

R-((-20R)(5R+70))=0

We multiply parentheses ..

-((-100R^2-1400R))+R=0


We calculate terms in parentheses: -((-100R^2-1400R)), so:

(-100R^2-1400R)

We get rid of parentheses

-100R^2-1400R

Back to the equation:

-(-100R^2-1400R)


We get rid of parentheses

100R^2+1400R+R=0

We add all the numbers together, and all the variables

100R^2+1401R=0

a = 100; b = 1401; c = 0;
Δ = b2-4ac
Δ = 14012-4·100·0
Δ = 1962801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1962801}=1401$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1401)-1401}{2*100}=\frac{-2802}{200}
=-14+1/100
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1401)+1401}{2*100}=\frac{0}{200}
=0
$