Algebra Question

Question

A rectangular piece of metal is 10in longer than it is wide. Squares with sides 2in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 598in^3​, what were the original dimensions of the piece of​ metal?

Answer 1

The rectangle has length w+10 and width w

The box as length w+10-2-2 = w+6, width w-4 and height 2

Volume = length*width*height

598 = (w+6)(w-4)*2

299 = w^2 + 2w - 24

0 = w^2 + 2w - 323

0 = (w + 19)(w - 17)

width w=17

the rectangle is 27 by 17...

the box is 23 by 13 by 2