document.write( "Question 1117918: The length of a rectangle is one less than twice the width. The area of the rectangle is 66. What are the dimensions of the rectangle? Only an algebraic solution will be accepted. \n" ); document.write( "
Algebra.Com's Answer #734306 by Shin123(626)\"\" \"About 
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So \"%282x-1%29%2Ax=66\" 2x-1 is the length and x is the width. Turn that to a quadratic equation,\r
\n" ); document.write( "\n" ); document.write( "\"2x%5E2-x-66=0\" Use the quadratic formula.\r
\n" ); document.write( "\n" ); document.write( "\"x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a\"\r
\n" ); document.write( "\n" ); document.write( "\"x=%28-%28-1%29%2B-sqrt%28-1%5E2-4%2A2%2A-66%29%29%2F2%282%29\"\r
\n" ); document.write( "\n" ); document.write( "\"x=%281%2B-sqrt%281-8%2A-66%29%29%2F4\"\r
\n" ); document.write( "\n" ); document.write( "\"x=%281%2B-sqrt%28529%29%29%2F4\"\r
\n" ); document.write( "\n" ); document.write( "\"x=%281%2B-+23%29%2F4\" Ignore the minus because we're looking for positive answers.\r
\n" ); document.write( "\n" ); document.write( "\"highlight%28x=6%29\" The width is \"66%2F6=highlight%2811%29\"
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