document.write( "Question 1156399: If the length of a rectangle is 7 meters less than twice the width, we have
\n" ); document.write( "L=2W-7
\n" ); document.write( "If the area is 60 square meters, we have
\n" ); document.write( "L×W=60 substitute L
\n" ); document.write( "(2W-7)×W=60
\n" ); document.write( "2W^2-7W=60
\n" ); document.write( "2W^2-7W-60=0
\n" ); document.write( "2W^2-15W+8W-60=0
\n" ); document.write( "(2W^2+8W)-(15W-60)=0
\n" ); document.write( "2W(W+4)-15(W+4)=0
\n" ); document.write( "(W+4)(2W-15)=0
\n" ); document.write( "2W-15=0 W=15/2 W=7.5
\n" ); document.write( "Can someone explain where the 15 and 8 came from?
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Algebra.Com's Answer #779101 by ikleyn(52788)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "May I give an advise to you ?\r\n" );
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document.write( "The problem is reduced to the quadratic equation \r\n" );
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document.write( "    2W^2 - 7W - 60 = 0.\r\n" );
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document.write( "From this point, simply solve it using the quadratic formula.\r\n" );
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document.write( "That is all.\r\n" );
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