document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=355495>Question 355495</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, I'm being asked to solve for x by completing the square:\r<BR>\n" );
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document.write( "-2x^2 + 12x - 6 = 0\r<BR>\n" );
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document.write( "-2x^2 + 12x = 6        < I moved the 6 to the other side\r<BR>\n" );
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document.write( "x^2 - 6x = -3          < I divided everything by -2 to get the x^2 by itself\r<BR>\n" );
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document.write( "x^2 - 6x + 9 = 6       < I took half of the coefficient 6 (6/2 = -3), squared <BR>\n" );
document.write( "                         it (9), and added it to both sides of the equation \r<BR>\n" );
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document.write( "This is where I get stuck. I'm supposed to write the left side in squared form, but because there is a plus and minus, it wouldn't be just one factor squared. Would I write this as (x-3)(x+3)? If so, how would I take the square root of that? \r<BR>\n" );
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document.write( "Thanks for any guidance you can provide! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #253889 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=253889\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> (x - 3)^2 = 6\r<BR>\n" );
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document.write( "x - 3 = ± sqrt(6)\r<BR>\n" );
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document.write( "x = 3 ± sqrt(6) </SPAN>\n" );
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