document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=499367>Question 499367</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The width of a rectangle is 2 less than twice its length. If the area of the rectangle is 53 cm squared, what is the length of the diagonal? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #337629 by </B><A HREF=/tutors/aboutme.mpl?userid=chessace><B>chessace(471)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=chessace><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/cgi-bin/show-face.mpl?user=chessace><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=337629\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> W = 2*L-2 (cm I assume)<BR>\n" );
document.write( "A = L*W = 53 cm^2<BR>\n" );
document.write( "A = 2L^2 -2L<BR>\n" );
document.write( "2L^2 -2L -53 = 0<BR>\n" );
document.write( "L = (2 +- sqrt(4+2*53))/2 = 1 + sqrt(110)/2 = 6.244<BR>\n" );
document.write( "W = 10.488<BR>\n" );
document.write( "Diag^2 = L^2 + W^2 = 39 + 110 = 149<BR>\n" );
document.write( "D = 12.207 cm<BR>\n" );
document.write( "Looks like a typo in the problem.<BR>\n" );
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