# Math

posted by
**Anonymous** on
.

An open box contains 80cm^3 and is made from a square piece of tinplate with 3cm squares cut from each of its four corners. Find the dimensions of the original piece of tinplate.

A=80

A=(length)(width)(height)

since length and width are the same, and when the box is folded the height will be 3cm,

80=(w^2)(3)

Since the width has 2 3cm sections cut out of it, I said:

w=x+6

To find x, I tried subbing w=x+6 into my previous equation:

80=((x+6)^2)*3

Which expands as:

(3x^2)+(36x)+28=0

What do I do next? I tried the quadratic formula to get x, but ended up with two negative numbers, which is not a valid solution. Please help me... Was my process wrong?