R(-4)=(x+1)(x-3)

Solution for R(-4)=(x+1)(x-3) equation:

(-4)=(R+1)(R-3)

We move all terms to the left:

(-4)-((R+1)(R-3))=0

We add all the numbers together, and all the variables

-((R+1)(R-3))-4=0

We multiply parentheses ..

-((+R^2-3R+R-3))-4=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+3-4=0

We add all the numbers together, and all the variables

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

a = -1; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·(-1)·(-1)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$R=\frac{-b}{2a}=\frac{-2}{-2}=1$