Math- Algebra

This one showed up in the middle of my quadratic equations. There is no y or 0. What am I supposed to do with it?

Give exact and approximate solutions to three decimal places.

(x+5)^2=81

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  1. x + 5 = + sqrt 81 = +9
    or
    x + 5 = - sqrt 81 = -9

    so
    x = 9-5 = 4
    or
    x = -9-5 = -14
    check those back

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  2. How to check back:
    if x = 4
    what is (x+5)^2 ? hint 9^2 = 81 !!!

    if x = -9
    what is (x+5)^2 ? hint (-9)*(-9) = 81
    we did it :)

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  3. perhaps it is coffee time

    if x = -14
    what is (x+5)^2 ? hint (-9)*(-9) = 81
    we did it :)

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  4. Oh, and to do this as quadratic equation:
    (x+5)^2 = 81
    x^2 + 10 x + 25 = 81

    x^2 + 10 x - 56 = 0

    x = -10/2 +/- (1/2) sqrt(100 + 4*1*56)

    x = -5 +/- (1/2)sqrt ( 324)

    sqrt 324 = 18 by the way

    x = -5 +/- 9

    x = -5+9 = +4
    or
    x = -5-9 = -14

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  5. You catching on to this ?

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  6. LOL. Your great, thanks.

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  7. Yeah, well check the one below. I did not read carefully and left the wall of the barn out !!

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