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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+R^2-3R+R-3))+R=0


We calculate terms in parentheses: -((+R^2-3R+R-3)), so:

(+R^2-3R+R-3)

We get rid of parentheses

R^2-3R+R-3

We add all the numbers together, and all the variables

R^2-2R-3

Back to the equation:

-(R^2-2R-3)


We add all the numbers together, and all the variables

R-(R^2-2R-3)=0

We get rid of parentheses

-R^2+R+2R+3=0

We add all the numbers together, and all the variables

-1R^2+3R+3=0

a = -1; b = 3; c = +3;
Δ = b2-4ac
Δ = 32-4·(-1)·3
Δ = 21
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{-(3)-\sqrt{21}}{2*-1}=\frac{-3-\sqrt{21}}{-2}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{21}}{2*-1}=\frac{-3+\sqrt{21}}{-2}
$