document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=48853>Question 48853</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help with this problem.  I have been doing fine with consecutive integer problems until the \"squares\" part was added.\r<BR>\n" );
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document.write( "Q: The sum of the squares of two consecutive odd integers decreased by the product of the integers is the same as 67.  \r<BR>\n" );
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document.write( "I have been setting it up like this, but think I may be starting wrong since I don't seem to get answers that make sense\r<BR>\n" );
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document.write( " x^2+(x+2)^2 - x(x+2) = 67 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #32349 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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=32349\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> your formula is correct x~2+(x+2)~2-x(x+2)=67 or x~2+x~2+4x+4-x~2-2x=67 or\r<BR>\n" );
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document.write( "x~2-2x-63=0 or (x-7)(x-9)=0 or x=7 & x=9 \r<BR>\n" );
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document.write( "proof 7~2+9~2-7*9=67 or 49+81-63=67 or 130-63=67 or 67=67  </SPAN>\n" );
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