document.write( "Question 145133: solve the following using the quadratic formula\r
\n" ); document.write( "\n" ); document.write( "x^2+6x+1=0\r
\n" ); document.write( "\n" ); document.write( "4x^2-12x+9=0
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Algebra.Com's Answer #105819 by solver91311(24713)\"\" \"About 
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Both of your equations are in the form \"ax%5E2%2Bbx%2Bc=0\". For example, for the first one: \"a=1\", \"b=6\", and \"c=1\".\r
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\n" ); document.write( "\n" ); document.write( "Just plug the values for \"a\", \"b\", and \"c\" into:\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\" \r
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\n" ); document.write( "\n" ); document.write( "and then do the arithmetic. Notice the 'plus or minus' sign, meaning that you will have two solutions for each equation.\r
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\n" ); document.write( "\n" ); document.write( "Also, check the number under the radical for perfect square factors, and take them out. For example \"sqrt%288%29=sqrt%284%2A2%29=sqrt%284%29sqrt%282%29=2sqrt%282%29\"\r
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\n" ); document.write( "\n" ); document.write( "Also, note that the coefficient on the 1st degree term in your second equation includes the minus sign.
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