Arectangular piece of metal is 10 in. longer than it is wide. squares with sides 2 in. 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 832 in. cubed, what were the original dimensions of the piece of metal

Volume of the box = length  * width * height

length of original peice = x + 10 , width = x

length of box  (x + 10) - 4  = x + 6
width of box = x - 4
height of box =  2

so we have 

2 * (x -4) * ( x + 6) = 832

2x^2 + 4x - 48 = 832

2x^2 + 4x - 880 = 0
x^2  + 2x - 440 = 0

(x + 22)(x - 20) = 0

x = 20

so width of original piece of metal = 20 ins , length = 30 inches.

That is right sorry i was gonna answer but i already saw his