Assuming the is [tex]A = s^2[/tex] you would need to take the square root of both sides. Normally, we insert +or- when using square root to solve.
Considering this formula is for the area of a square and sides are a length and can't be represented by a negative number we would not need the -.
Therefore
[tex]s= \sqrt{A} [/tex]
but I would speak with my instructor about the issue.
