a rectangular piece of metal is 15in longer then it is wide. squares with sides 3in 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 1092in cubed what were the original dimensions of the piece of metal?

Step-by-step explanation:

Let x equal the width of the box and x + 15 equal the length of the box.  Subtract 6 (the amount cut out) from both of the these to get the following terms: (x+9) and (x-6).  These are the length and width of the box respectively.  We also know that the height of the box is 3.  Plugging these into the formula for the volume of a box gives you:

3(x+9)(x-6)=1092

Expanding this out gives:

3x^2+9x-162=1092

Subtract 1092 from both sides to make the quadratic equation equal zero.  Plug the a, b and c coefficients into the quadratic formula (or factor) to find x.

3x^2 +9x-1254 = 0

a = 3       b = 9        c = -1254

I hope this helps :)

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629

c

step-by-step explanation: