document.write( "Question 1057459: Find three consecutive negative integers such that the product of the first two is 26 greater than the square of the third. \r
\n" ); document.write( "\n" ); document.write( "For some reason I end up with a positive integer after working it out, any help? \n" ); document.write( "
Algebra.Com's Answer #672517 by ankor@dixie-net.com(22740)\"\" \"About 
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Find three consecutive negative integers such that the product of the first two is 26 greater than the square of the third.
\n" ); document.write( "let our numbers be:(n-2), (n-1), n
\n" ); document.write( "(n-2)(n-1) = n^2 + 26
\n" ); document.write( "FOIL the left
\n" ); document.write( "n^2 - 3n + 2 = n^2 + 26
\n" ); document.write( "combine like terms
\n" ); document.write( "n^2 - n^2 - 3n = 26 - 2
\n" ); document.write( "-3n = 24
\n" ); document.write( "n = 24/-3
\n" ); document.write( "n = -8
\n" ); document.write( ":
\n" ); document.write( "the three number -10, -9, -8
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "see if the works
\n" ); document.write( "-10 *-9 = -8^2 + 26
\n" ); document.write( "+90 = +64 + 26 \n" ); document.write( "
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