document.write( "Question 89065: solve:\r
\n" ); document.write( "\n" ); document.write( "(x+2)^2=3
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Algebra.Com's Answer #64741 by malakumar_kos@yahoo.com(315)\"\" \"About 
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\n" ); document.write( " solve:\r
\n" ); document.write( "\n" ); document.write( "(x+2)^2=3\r
\n" ); document.write( "\n" ); document.write( " consider (x+2)^2 = 3
\n" ); document.write( " x^2+4x+4-3= 0
\n" ); document.write( " x^2+4x+1 = 0\r
\n" ); document.write( "\n" ); document.write( " use the formula x = -b +-sq rt b^2-4ac/2a where a = 1 b = 4 c = 1
\n" ); document.write( " x = -1+-sq rt 1^2-4.1.1/2.1
\n" ); document.write( " = -1+-sq rt 1-4/2
\n" ); document.write( " = -1+sq rt-3/2 since sq rt of negative number cannot be found we consider the solution as x = (-1+-sqrt-3)/2\r
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