document.write( "Question 65770: Solve the quadratic equation by completing the square:
\n" ); document.write( "2x^2 + 10x + 11 = 0\r
\n" ); document.write( "\n" ); document.write( "Can someone please explain to me the idea behind the quadratic equation, thank you
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Algebra.Com's Answer #46495 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
A quadratic equation is one in which the highest power of the independent variable (x in this case) is 2. Quadratic comes from the latin \"Quadra\" meanining square.\r
\n" ); document.write( "\n" ); document.write( "Solve by completing the square:
\n" ); document.write( "\"2x%2A2+%2B10x+%2B+11+=+0\" First,divide through by 2 in order to get the coefficient of \"x%5E2\" = 1.
\n" ); document.write( "\"x%5E2+%2B+5x+%2B+11%2F2+=+0\" Next, subtract 11/2 from both sides.
\n" ); document.write( "\"x%5E2+%2B+5x+=+-11%2F2\" Now add to both sides, the square of half the coefficient of x. This is:\"%285%2F2%29%5E2+=+25%2F4\"
\n" ); document.write( "\"x%5E2+%2B+5x+%2B+25%2F4+=+%2825%2F4%29-11%2F2\" Now simplify this.
\n" ); document.write( "\"%28x%2B5%2F2%29%5E2+=+3%2F4\" Take the square root of both sides.
\n" ); document.write( "\"x%2B5%2F2+=+sqrt%283%2F4%29\" The \"sqrt%283%2F4%29\" will have a + or - in front of it. Subtract \"5%2F2\" from both sides.
\n" ); document.write( "\"x+=+%28-5%2F2%29%2B-sqrt%283%2F4%29\"\r
\n" ); document.write( "\n" ); document.write( "The solutions are:
\n" ); document.write( "\"x+=+%28-5%2Bsqrt%283%29%29%2F2\" = -1.63397 (Approx.)
\n" ); document.write( "\"x+=+%28-5-sqrt%283%29%29%2F2\" = -3.36603 (Approx.)\r
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