document.write( "Question 892730: The length of a rectangle is 3cm more than its width it has an area of 475cm (squared)
\n" ); document.write( "what is the width to 1 decimal place? \r
\n" ); document.write( "\n" ); document.write( "Please help! :) thank you
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Algebra.Com's Answer #540742 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
For a rectangle,
\n" ); document.write( "\"A=L%2AW\"
\n" ); document.write( "In this case,
\n" ); document.write( "\"L%2AW=475\"
\n" ); document.write( "and
\n" ); document.write( "\"L=3%2BW\"
\n" ); document.write( "Substituting,
\n" ); document.write( "\"%283%2BW%29W=475\"
\n" ); document.write( "\"W%5E2%2B3W=475\"
\n" ); document.write( "\"W%5E2%2B3W%2B%289%2F4%29=475%2B9%2F4\"
\n" ); document.write( "\"%28W%2B3%2F2%29%5E2=1900%2F4%2B9%2F4\"
\n" ); document.write( "\"%28W%2B3%2F2%29%5E2=1909%2F4\"
\n" ); document.write( "\"W%2B3%2F2=0+%2B-+sqrt%281909%29%2F2\"
\n" ); document.write( "\"W=-3%2F2+%2B-+sqrt%281909%29%2F2\"
\n" ); document.write( "Only the positive solution makes sense in this problem.
\n" ); document.write( "\"highlight%28W=sqrt%281909%29%2F2-3%2F2%29\"
\n" ); document.write( "Then,
\n" ); document.write( "\"L=3%2Bsqrt%281909%29%2F2-3%2F2\"
\n" ); document.write( "\"highlight%28L=sqrt%281909%29%2F2%2B3%2F2%29\"
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