The equilibrium concentration of #D# is 0.890 M.
The first part of the problem was already answered here, check it out:
http://socratic.org/questions/initially-only-a-and-b-are-present-each-at-2-00-m-what-is-the-final-concentratio?source=search
Now for the second part. This time, you start with 1.00 M of #A# and 2.00 M of #B#. Once again, use an ICE table to determine the equilibrium concentration of #D#.
#" "A " "+ " "B " "rightleftharpoons " "C " "+" " D#
I....1.00.............2.00...............0...................0
C....(-x)...............(-x)...............(+x)...............(+x)
E...1.00-x..........2.00-x...........x....................x
#K_c = ([C] * [D])/([A] * [B]) = (x * x)/((1.00-x)*(2.00-x))#
#K_c = x^2/((1.00-x)(2.00-x)) = 6.5#
Your cuadratic equation will look like this
#5.5x^2 - 19.5x + 13 = 0#
This equation will produce two solutions for #x#, 2.66 and 0.890. The first solution will be eliminated, since it implies that the equilibrium concentrations of #A# and #B# are negative.
This means that the equilibrium concentration of #D# will be
#[D] = color(green)("0.890 M")#
