document.write( "Question 1170681: The area of a rectangle is 53m^2 and the length of the rectangle is 5 m less than twice the width, find the dimensions of the rectangle
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Algebra.Com's Answer #795573 by MathLover1(20850)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "The area of a rectangle is \"53m%5E2\": \r
\n" ); document.write( "\n" ); document.write( "\"L%2AW=53\"\r
\n" ); document.write( "\n" ); document.write( "and if the length \"L\" of the rectangle is \"5m\" less than \"twice\" the width \"W\", we have \r
\n" ); document.write( "\n" ); document.write( "\"L=2W-5\"....substitute in area\r
\n" ); document.write( "\n" ); document.write( "\"%282W-5%29%2AW=53\"\r
\n" ); document.write( "\n" ); document.write( "\"2W%5E2-5W=53\"\r
\n" ); document.write( "\n" ); document.write( "\"2W%5E2-5W-53=0\"\r
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\n" ); document.write( "\n" ); document.write( "\"W+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\" \r
\n" ); document.write( "\n" ); document.write( "\"W+=+%28-%28-5%29+%2B-+sqrt%28+%28-5%29%5E2-4%2A2%2A%28-53%29+%29%29%2F%282%2A2%29+\" \r
\n" ); document.write( "\n" ); document.write( "\"W+=+%285+%2B-+sqrt%28+25%2B424+%29%29%2F4+\" \r
\n" ); document.write( "\n" ); document.write( "\"W+=+%285+%2B-+sqrt%28+449+%29%29%2F4+\" \r
\n" ); document.write( "\n" ); document.write( "\"W+=+%285+%2B-+21.19%29%2F4+\" \r
\n" ); document.write( "\n" ); document.write( "solutions: need only positive\r
\n" ); document.write( "\n" ); document.write( "\"W+=+%285+%2B+21.19%29%2F4+\" \r
\n" ); document.write( "\n" ); document.write( "\"W+=+6.55\" \r
\n" ); document.write( "\n" ); document.write( "go to\r
\n" ); document.write( "\n" ); document.write( "\"L=2W-5\" substitute \"W\"\r
\n" ); document.write( "\n" ); document.write( "\"L=2%2A6.55-5\"\r
\n" ); document.write( "\n" ); document.write( "\"L=8.1\"\r
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\n" ); document.write( "\n" ); document.write( " the dimensions of the rectangle are approximately \"8.1\" by \"+6.55\" \r
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