document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=570028>Question 570028</A>: <I><SPAN STYLE=\"word-break: break-all;\"> x2-2x=48....i have to solve for x by completing the square..how do you do this?<BR>\n" );
document.write( " ^(squared) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #367509 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=367509\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "<i><b>Step 1.</b></i>  Move the constant term to the right hand side.  This step is already done.\r<BR>\n" );
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document.write( "<i><b>Step 2.</b></i>  Divide through by the lead coefficient.  The lead coefficient in this problem is 1, so you can skip this step.\r<BR>\n" );
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document.write( "<i><b>Step 3.</b></i>  Divide the coefficient on the first degree term by 2.  In this case, 2 divided by 2 is 1.\r<BR>\n" );
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document.write( "<i><b>Step 4.</b></i> Square the result of Step 3.  1 squared is 1.\r<BR>\n" );
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document.write( "<i><b>Step 5.</b></i> Add the result of Step 4 to both sides of the equation.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ +\ 1\ =\ 49\">\r<BR>\n" );
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document.write( "<i><b>Step 6.</b></i> Factor the perfect square trinomial in the LHS:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ (x\ + 1)^2\ =\ 49\">\r<BR>\n" );
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document.write( "<i><b>Step 7.</b></i> Take the square root of both sides, considering both the positive and negative root:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ \pm\sqrt{49}\ =\ \pm7\">\r<BR>\n" );
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document.write( "So\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ 7\">\r<BR>\n" );
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document.write( "or\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ -7\">\r<BR>\n" );
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document.write( "<i><b>Step 8.</b></i> Solve the two linear equations to get your two roots.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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