document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=831456>Question 831456</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the length is two more than two times the width. if the area is 84cm^2 what are the dimesions of the rectangle </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #501389 by </B><A HREF=/tutors/aboutme.mpl?userid=hovuquocan1997><B>hovuquocan1997(83)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=hovuquocan1997><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=501389\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the width be y in the equation<BR>\n" );
document.write( "Two times the width is 2y, and two more than that is 2y + 2<BR>\n" );
document.write( "So you have the length is (2y+2) and the width is just y (as we let it be)<BR>\n" );
document.write( "Area is length times width, which is (2y+2)*y = 84<BR>\n" );
document.write( "2y^2 + 2y - 84 = 0<BR>\n" );
document.write( "Factor and solve<BR>\n" );
document.write( "(2y - 12)*(y + 7) = 0<BR>\n" );
document.write( "y = 6 or y = -7 <BR>\n" );
document.write( "But of course the width can't be a negative number so y = 6 is the answer<BR>\n" );
document.write( "Now find the length : 2y+2 = 2*6 + 2 = 14<BR>\n" );
document.write( "TA-DAH<BR>\n" );
document.write( "The length is 14 and the width is 6 :D  </SPAN>\n" );
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