document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=39406>Question 39406</A>: <I><SPAN STYLE=\"word-break: break-all;\">  Need help on solving these two problems, thanks =0) <BR>\n" );
document.write( "1. -9x+20=-2x<BR>\n" );
document.write( "2.  2x^2-5=3x  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #24851 by </B><A HREF=/tutors/aboutme.mpl?userid=corrie21><B>corrie21(4)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=corrie21><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=24851\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1.  -9x+20=-2x\r<BR>\n" );
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document.write( "First, add 9x to -9x, these cancel out.  Then add 9x to the -2x.\r<BR>\n" );
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document.write( "20=7x\r<BR>\n" );
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document.write( "to get the x by itself, divide the 7 into the 20.\r<BR>\n" );
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document.write( "20/7=x\r<BR>\n" );
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document.write( "2.  2x^2-5=3x\r<BR>\n" );
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document.write( "First, minus the 3x from both sides.<BR>\n" );
document.write( "2x^2-3x-5=0\r<BR>\n" );
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document.write( "then factor out\r<BR>\n" );
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document.write( "(2x-5)(x+1) </SPAN>\n" );
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