document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=135484>Question 135484</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The lenght of a rectangular yard is 25ft more than its width. If the width were doubled, the area of the yard would be 1800ft2. Find the dimensions of the yard. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #99299 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley77><B>checkley77(12844)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley77><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=99299\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> L=W+25<BR>\n" );
document.write( "L*2W=1800<BR>\n" );
document.write( "(W+25)2W=1800<BR>\n" );
document.write( "2W^2+50W-1800=0<BR>\n" );
document.write( "2(W^2+25W-900)=0<BR>\n" );
document.write( "2(W-20)(W+45)=0<BR>\n" );
document.write( "W-20=0<BR>\n" );
document.write( "W=20 FOR THE ORIGINAL WIDTH.<BR>\n" );
document.write( "L=20+25=45 FOR THE ORIGINAL LENGTH.<BR>\n" );
document.write( "20*2(20+25)=1800<BR>\n" );
document.write( "20*2*45)=1800<BR>\n" );
document.write( "1800=1800 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );