MATH PROBLEM HELP!!!!!!!!!

A rectangle anda square have the same area. the width of the rectangle is 96cm less than the side of the square and the length of the rectangle is 80 cm more than twice the side of the square. what are the dimensions of the rectangle

let your square have a side length of x cm
then your rectangle width is (x-96) and the length is (2x+80)

Area of square is x^2
Area of rectangel is (x-96)(2x+80)

Hence as the area for both are equal
x^2 = (x-96)(2x+80)
x^2 = 2x^2 - 192x + 80x - 7680
simplfiying you will get
x^2 - 112x -7680 = 0
Solving for x

(to solve for ax2 + bx + c = 0, Use the link below )

You will get x = 160 or x = -48 (rejected cause x cannot be negative)
Hence the dimensions of your rectangle is width 160 - 96 = 64cm, length is 2x160 + 80 = 400cm

Let the areas=A
Width of the rectangle=w
Lenght of the rectancle = L
And side of square= s

From area formulas you know the area of a square = s*s=A
And the area of a rectangle = w*L=A

So s*s=w*L because the areas are the same.

Now you need 2 more equations because we have 3 variables to solve for.

Since the width or the rectangle is 96 less than the side of the square we have
W-96=s

And the lenght of the rectangle is 80 more than twice the side of the square
L+80=2*s

So your system of equations is:
S*s=w*L
W-96=s
L+80=2*s

Now you solve for w and L and since I'm on my phone ill leave that to you. Good luck hope that helps!

Rectangle of L and W

Square of X

W = X - 96

L = 2X + 80

LW = X^2

(X - 96)(2X + 80) = X^2

2X^2 - 192 X + 80X - 7680 - X^2 = 0

X^2 - 112X - 7680 = 0

X = 160

L = 2X + 80 = 400
W = X - 96 = 64