document.write( "Question 1116041: The length of a rectangle is 5 ft
\n" ); document.write( " more than double the width, and the area of the rectangle is 42 ft^2
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Algebra.Com's Answer #730869 by rothauserc(4718)\"\" \"About 
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let l be the length of the rectangle and w be its width
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\n" ); document.write( "l = 2w +5
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\n" ); document.write( "l * w = area of rectangle
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\n" ); document.write( "(2w+5) * w = 42
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\n" ); document.write( "2w^2 +5w = 42
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\n" ); document.write( "subtract 42 from both side of the =
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\n" ); document.write( "2w^2 +5w -42 = 0
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\n" ); document.write( "(2w-7) * (w+6) = 0
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\n" ); document.write( "there are two solutions for w
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\n" ); document.write( "w = 7/2 = 3.5 and w = -6
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\n" ); document.write( "we reject the negative value for w
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\n" ); document.write( "l = 2(3.5) +5 = 12
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\n" ); document.write( "l = 12 feet and w = 3.5 feet
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