document.write( "Question 629291: Find three consecutive odd integers such that the product of the first two minus six times the third is 21. \n" ); document.write( "

Algebra.Com's Answer #397466 by dfrazzetto(283) $\"\"$ $\"About$

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2n+1, 2n+3, 2n+5\r
\n" ); document.write( "\n" ); document.write( "(2n+1)(2n+3) - 6(2n+5) = 21\r
\n" ); document.write( "\n" ); document.write( "4n^2 + 8n + 3 - 12n - 30 = 21\r
\n" ); document.write( "\n" ); document.write( "4n^2 -4n -27 -21 = 0\r
\n" ); document.write( "\n" ); document.write( "4n^2 - 4n - 48 = 0\r
\n" ); document.write( "\n" ); document.write( "n^2 - n - 12 = 0\r
\n" ); document.write( "\n" ); document.write( "(n+3)(n-4) = 0\r
\n" ); document.write( "\n" ); document.write( "n = -3, 4\r
\n" ); document.write( "\n" ); document.write( "two sets of numbers satisfy: \r
\n" ); document.write( "\n" ); document.write( "{-5, -3, -1}
\n" ); document.write( "{9, 11, 13}\r
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