document.write( "Question 782249: Solve using the square root method.
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Algebra.Com's Answer #476339 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "\"%282x-3%29%5E2=9\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"2x-3=%22%22%2B-sqrt%289%29\" Take the square root of both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"2x-3=sqrt%289%29\" or \"2x-3=-sqrt%289%29\" Break up the \"plus/minus\" to form two equations.\r
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\n" ); document.write( "\n" ); document.write( "\"2x-3=3\" or \"2x-3=-3\" Take the square root of \"9\" to get \"3\".\r
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\n" ); document.write( "\n" ); document.write( "\"2x=3%2B3\" or \"2x=3-3\" Add \"3\" to both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"2x=6\" or \"2x=0\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"x=3\" or \"x=0\" Divide both sides by \"2\" to isolate \"x\".\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So the solutions are \"x=3\" or \"x=0\".
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