document.write( "Question 66194This question is from textbook merrill algebra two with trigonometry
\n" ); document.write( ": If the product of two consecutive odd integers is decreased by one-third the lesser integer, the result is 250. Find the integers.
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Algebra.Com's Answer #46941 by checkley71(8403)\"\" \"About 
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x(x+2)-x/3=250
\n" ); document.write( "(x^2+2x)-x/3=250
\n" ); document.write( "(3x^2+6x-x)/3=250
\n" ); document.write( "3x^2+5x=750
\n" ); document.write( "3x^2+5x-750=0 using the quadratic equation we get
\n" ); document.write( "x=(-5+-sqrt[5^5-4*3*-750])/2*3
\n" ); document.write( "x=(-5+-sqrt[25+9000])/6
\n" ); document.write( "x=(-5+-sqrt9025])/6
\n" ); document.write( "x=(-5+-95)/6
\n" ); document.write( "x=(-5+95)/6
\n" ); document.write( "x=90/6
\n" ); document.write( "x=15 solution
\n" ); document.write( "x=(-5-95)/6
\n" ); document.write( "x=-100/6
\n" ); document.write( "x=-16 2/3
\n" ); document.write( "proof
\n" ); document.write( "15(15+2)-15/3=250
\n" ); document.write( "15*17-5=250
\n" ); document.write( "255-5=250
\n" ); document.write( "250=250\r
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