# How do I answer this question? `(n-5)^2=32`

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### 3 Answers

`(n-5)^2 = 32`

`(n-5) = +-sqrt32`

`n = 5+-sqrt32`

`n = 5+-sqrt(16xx2)`

`n =5+-sqrt16xxsqrt2`

`n = 5+-4sqrt2`

*So the answers are;*

`n = 5+4sqrt2`

`n = 5-4sqrt2`

`(n-5)^2=32`

To solve, expand left side.

`(n-5)(n-5)=32`

`n^2-10n+25=32`

Then, set one side equal to zero.

`n^2-10n+25-32=32-32`

`n^2-10n-7=0`

And, use the quadratic formula to solve for n.

`n=(-b+-sqrt(b^2-4ac))/(2a)=(-(-10)+-sqrt((-10)^2-4*1*(-7)))/(2*1) `

`n=(10+-sqrt(100+28))/2=(10+-sqrt128)/2=(10+-8sqrt2)/2=5+-4sqrt2`

**Hence, `n=5+4sqrt2` and `n=5-4sqrt2` .**

they have already done things halfway for you the next steps are very brief and fast

`n-5=+-sqrt(32)`

`n=5+-sqrt(32)`

so n is both

`5-sqrt(32)`

`5+sqrt(32)`