How would I extract this square root?

Question

How would I extract this...

(r+6)^2 = 13

Answer 1

(r+6)^2 = 13
r^2 + 12x + 36 = 13
r^2 + 12x + 23 = 0

Plug into quadratic equation:

[-b +- sqrt (b^2 - 4ac)]/2a
where a is the r^2 coefficient,
b is the r coefficient
c is constant.

Or
[-12 +- sqrt(12^2 - 4*23)]/2
[-12 +- sqrt(144 - 92)]/2
[-12 +- sqrt(54)]/2
[-12 +- sqrt(2*3*3*3)]/2
[-12 +- 3 sqrt(6)]/2

Edit: Maybe it would help if people understood that (r+6)^2 was the same as (r+6)(r+6). You have to apply FOIL method of polynomial multiplication. You also can't get rid of a positive 36 by adding 36 to both sides (that would give you a 72 on one side and 49 on the right side) or by subtracting from one side and adding to the other (that violates laws of algebra that require you to do the same thing to both sides of the equation).

Answer 2

First multiply the term in parentheses by itself ("square it") to get r^2 + 36 = 13. Then combine the numbers and get r^2 = 49.

Now take the square root of 49 which is7 (positive and negative).

That makes (x+7)(x-7) = 0 so x = +7 and - 7

Answer 3

r^2+36=13

Answer 4

First dig around, and then pull really hard, it should come out.