R(11)=(x+1)(x-3)

Solution for R(11)=(x+1)(x-3) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

-((+R^2-3R+R-3))+11=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 get rid of parentheses

-R^2+2R+3+11=0

We add all the numbers together, and all the variables

-1R^2+2R+14=0

a = -1; b = 2; c = +14;
Δ = b2-4ac
Δ = 22-4·(-1)·14
Δ = 60
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{15}}{2*-1}=\frac{-2-2\sqrt{15}}{-2}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{15}}{2*-1}=\frac{-2+2\sqrt{15}}{-2}
$