0=(2r+1)(2r+1)

Solution for 0=(2r+1)(2r+1) equation:

0=(2r+1)(2r+1)

We move all terms to the left:

0-((2r+1)(2r+1))=0

We add all the numbers together, and all the variables

-((2r+1)(2r+1))=0

We multiply parentheses ..

-((+4r^2+2r+2r+1))=0


We calculate terms in parentheses: -((+4r^2+2r+2r+1)), so:

(+4r^2+2r+2r+1)

We get rid of parentheses

4r^2+2r+2r+1

We add all the numbers together, and all the variables

4r^2+4r+1

Back to the equation:

-(4r^2+4r+1)


We get rid of parentheses

-4r^2-4r-1=0

a = -4; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·(-4)·(-1)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$r=\frac{-b}{2a}=\frac{4}{-8}=-1/2$