document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=903029>Question 903029</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The area of a rectangle is 135 square meters. The width is four meters longer than 1/3 of the length. Find the length and the width..  I understand that the equation for width is w= 1/3l + 4 then I multiples the entire thing by l and set it equal to 135 which is the area. I got 135= 1/3l squared plus 4l. Then I started to factor and got 0= 1/3l^2 +4l -135. I just don't know where to go from there </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #547704 by </B><A HREF=/tutors/aboutme.mpl?userid=ewatrrr><B>ewatrrr(24785)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ewatrrr><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=ewatrrr><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=547704\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> L(L/3 - 4) = 135m^2<BR>\n" );
document.write( "L^2/3 - 4L = 135<BR>\n" );
document.write( "L^2 - 12L -405 = 0<BR>\n" );
document.write( "(L-27)(L+15)= 0  (Tossing out the negative solution for unit measure)<BR>\n" );
document.write( "L = 27m  and w = 5m<BR>\n" );
document.write( "And...<BR>\n" );
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