For a rectangle:A = Length x WidthL = W + 6A = ( W + 6 ) x WA = W² + 6 WW² + 6 W - 667 = 0W 1/2 = ( - 6 +/- √(6² - 4 · 1 · (-667)) / 2W 1/2 = ( - 6 +/- √ ( 36 + 2,668 ) ) / 2W = ( - 6 + √2,704 ) / 2 ( another solution is negative )W = ( - 6 + 52 ) / 2W = 46 / 2 W = 23 inches ( at least )If we want to prove it: W = 23, L = 23 + 6 = 2923 x 29 = 667
Answer: The possible widths are 23 inches and more.