document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=144660>Question 144660</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A rectangle has a diagonal that is 3.6 feet longer than the length and 7.1 feet longer than the width. What are the dimensions of the rectangle? I would appreaciate it if someone could help me solve this problem. This is what I've done so far but I know its not correct. Thank you for your time. <BR>\n" );
document.write( "X^2 + (X+7.1)^2 = (X+3.6)^2 <BR>\n" );
document.write( "X^2 + X^2 + 14.2 + 50.41 = X^2 + 2X + 3.6 <BR>\n" );
document.write( "X^2+16.2X + 52.6=0<BR>\n" );
document.write( "(X=13.15)(X=4.05)        </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #105432 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=105432\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let x=diagonal, so x-3.6=length and x-7.1=width\r<BR>\n" );
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document.write( "x^2=(x-3.6)^2+(x-7.1)^2 __ x^2=x^2-7.2x+12.96+x^2-14.2x+50.41\r<BR>\n" );
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document.write( "subtracting x^2 __ 0=x^2-21.4x+63.37\r<BR>\n" );
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document.write( "use quadratic formula to find x </SPAN>\n" );
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