Question 600301
{{{D=(n^2-3n)/2}}}



{{{2D=n^2-3n}}}



{{{0=n^2-3n-2D}}}



{{{n^2-3n-2D=0}}}



Now use the quadratic formula to solve for 'n'. In this case, {{{A=1}}}, {{{B = -3}}} and {{{C=-2D}}}



{{{n = (-B +- sqrt( B^2-4AC ))/(2A)}}}



{{{n = (-(-3) +- sqrt( (-3)^2-4(1)(-2D) ))/(2(1))}}}



{{{n = (3 +- sqrt( 9 + 8D ))/(2)}}}



{{{n = (3 + sqrt( 9 + 8D ))/(2)}}} or {{{n = (3 - sqrt( 9 + 8D ))/(2)}}}



So the solutions are {{{n = (3 + sqrt( 9 + 8D ))/(2)}}} or {{{n = (3 - sqrt( 9 + 8D ))/(2)}}}