document.write( "Question 154135: We are suppose to use the method that came from India to solve our problems, but I'm lost.\r
\n" ); document.write( "\n" ); document.write( "x^2-2x-13=0
\n" ); document.write( "x^2-2x-13+13=0
\n" ); document.write( "x^2-2x=13
\n" ); document.write( "4x^2-8x=52
\n" ); document.write( "4x^2-8x+4=52+4
\n" ); document.write( "4x^2-8x+4=56
\n" ); document.write( "the next step and there after I'm lost, could you please help me out
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Algebra.Com's Answer #113438 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Solve by \"the method that came from India\":
\n" ); document.write( "\"x%5E2-2x-13+=+0\"
\n" ); document.write( "Now I've never been to India, but the method to which you are refering is also known as \"completing the square\" and here's how you do that:
\n" ); document.write( "\"x%5E2-2x-13+=+0\" First, add 13 to both sides of the equation.
\n" ); document.write( "\"x%5E2-2x+=+13\" Now \"complete the square\" in x by adding the square of half the x-coefficient to both sides of the equation. This would be: \"%28-2%2F2%29%5E2+=+1\"
\n" ); document.write( "\"x%5E2-2x%2B1+=+13%2B1\" Simplify.
\n" ); document.write( "\"x%5E2-2x%2B1+=+14\" Now factor the left side of the equation.
\n" ); document.write( "\"%28x-1%29%28x-1%29+=+14\" Rewrite the left sides as:
\n" ); document.write( "\"%28x-1%29%5E2+=+14\" Now take the square root of both sides.
\n" ); document.write( "\"x-1+=+0%2B-sqrt%2814%29\" Finally, add 1 to both sides.
\n" ); document.write( "\"highlight%28x+=+1%2B-sqrt%2814%29%29\"
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