document.write( "Question 187686: Solve\r
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Algebra.Com's Answer #140719 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
I think you meant to write \"9x%5E2=14\" right?\r
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\n" ); document.write( "\n" ); document.write( "\"9x%5E2=14\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2=14%2F9\" Divide both sides by 9.\r
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\n" ); document.write( "\n" ); document.write( "\"x=sqrt%2814%2F9%29\" or \"x=-sqrt%2814%2F9%29\" Take the square root of both sides (don't forget the \"plus/minus\")\r
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\n" ); document.write( "\n" ); document.write( "\"x=sqrt%2814%29%2Fsqrt%289%29\" or \"x=-sqrt%2814%29%2Fsqrt%289%29\" Break up the square root.\r
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\n" ); document.write( "\n" ); document.write( "\"x=sqrt%2814%29%2F3\" or \"x=-sqrt%2814%29%2F3\" Take the square root of 9 to get 3\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=sqrt%2814%29%2F3\" or \"x=-sqrt%2814%29%2F3\"
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