document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=368683>Question 368683</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve by factoring.<BR>\n" );
document.write( "10t^2 - t - 2 = 0 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #262945 by </B><A HREF=/tutors/aboutme.mpl?userid=lefty4ever26><B>lefty4ever26(59)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lefty4ever26><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=lefty4ever26><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=262945\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The trick to factoring these kinds of trinomials is to take the 1st constant(10) multiplied by the last constant (-2).  This is -20.  What two numbers multiplied together equal -20 but add together to equal the middle constant of your equation(-1).  Well that would be -5 and 4.  because -5*4=-20 and -5+4=-1\r<BR>\n" );
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document.write( "So to factor it we replace the -t with (-5t+4t) to get\r<BR>\n" );
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document.write( "10t^2-5t+4t-2=0  Then we need to factor by grouping<BR>\n" );
document.write( "(10t^2-5t)+(4t-2)=0  <BR>\n" );
document.write( "5t(2t-1)+2(2t-1)=0  Then we combine to make the two factors<BR>\n" );
document.write( "(5t+2)(2t-1)=0  Set each factor equal to 0 and solve for t. <BR>\n" );
document.write( "5t+2=0   2t-1=0<BR>\n" );
document.write( "t=-2/5   t=1/2 \r<BR>\n" );
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document.write( "Hope this helps! </SPAN>\n" );
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