document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=175328>Question 175328</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I confused trying to solve 2x^2-5x-3>=0. i think you begin by adding 3 to both sides but then having the exponet is throwing me off. i dont know what to do after that. If you could please explain it would be much appreciated. thanks! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #130417 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=130417\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2x^2-5x-3>=0---------------------eq1<BR>\n" );
document.write( "(2x+1)(x-3)>=0<BR>\n" );
document.write( "Now our drill here is to determine the range of values for x that makes (2x+1)(x-3)>=0.  In other words (2x+1)(x-3) has to be either zero or (-)(-) or (+)(+)<BR>\n" );
document.write( "We can see by inspection that if x>=3, then<BR>\n" );
document.write( "(2x+1)(x-3) is either 0 or positive(+)(+) and eq1 inequality is correct<BR>\n" );
document.write( "We can also see by inspection that if x<=-1/2, then<BR>\n" );
document.write( "(2x+1)(x-3) is either 0 or positive(-)(-)and eq1 inequality is correct\r<BR>\n" );
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document.write( "So, our answer is:<BR>\n" );
document.write( "x>=3 or x<=-1/2\r<BR>\n" );
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document.write( "This is one of many ways to solve quadratic inequalities\r<BR>\n" );
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document.write( "Does this help?----ptaylor </SPAN>\n" );
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