Icouldn't figure this one out. pls? ?

29 a) Area = 400 in^2

length of the side = x

equation
400 in^2= (x)^2

It has two solutions, one positve and one negative

x =  +/- √[400] = +/- 20 in

x = 20 in and x = -20in

b) Only the postive solution, 20 in, makes sense given that the varialbe represents the length of the side of a true square and that only can be a positive number.

c) Find the perimeter = 4x = 4(20in) = 80 in.