# (N-5) squared =20 <--- I need help figuring this out. Not only the answer but maybe an explanation as to how?

(n - 5)^2 = 20

Expand the left side:
(n - 5)(n - 5) = 20
n^2 - 10n + 25 = 20

Now, set the equation equal to zero:
n^2 - 10n + 25 = 20
n^2 - 10n + 5 = 0

To solve for n, you'll have to use the quadratic formula:
n^2 - 10n + 5 = 0

[-b ± sqrt(b^2 - 4ac)] / 2a
Where a = 1, b = -10, c = 5

n = 5 - 2*sqrt(5) = 0.527864
n = 5 + 2*sqrt(5) = 9.47214

~ Mitch ~

P.S. - Ichweissallez's answer is the easiest to understand.
Put everything in the square route.

and once done, it should look like
When you put something in a sqrt you must make it [+/-]

(n-5)=[+/-]sqrt20
SIMPLIFY THE ROUTE
(n-5)=[+/-]2sqrt5
n=5[+/-]2sqrt5**

**edit when i did my last step I didn't copy correctly

people below me can't leave 20 in a sqrt because it's not simplified (2sqrt5 is more acceptable than sqrt20)
Here's how to start...

(n-5)(n-5) = 20

Use FOIL (first, outside, inside, last)
Solve for N

the guy's answer after mine makes a LOT more sense -- taking the square root of each side
You can take the square root of both sides of the equation, getting N-5 = +/- sqrt(20). Now you can add 5 to each side, getting N = 5 + sqrt(20) or N = 5 - sqrt(20). And that's all there is to it; numerical values, if needed, could be obtained with a calculator.

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