document.write( "Question 357480: 2x^2=200 \n" ); document.write( "
Algebra.Com's Answer #255253 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "\"2x%5E2=200\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2=%28200%29%2F%282%29\" Divide both sides by \"2\".\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2=100\" Reduce.\r
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\n" ); document.write( "\n" ); document.write( "\"x=%22%22%2B-sqrt%28100%29\" Take the square root of both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"x=sqrt%28100%29\" or \"x=-sqrt%28100%29\" Break up the \"plus/minus\" to form two equations.\r
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\n" ); document.write( "\n" ); document.write( "\"x=10\" or \"x=-10\" Take the square root of \"100\" to get \"10\".\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=10\" or \"x=-10\".
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