What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ?

You know that this reaction

#"A " + " B " rightleftharpoons " C " + " D"#

consumes #"A"# and #"B"# in a #1:1# mole ratio and produces #"C"# and #"D"# in a #1:1# mole ratio.

Moreover, you know that the equilibrium constant that describes this equilibrium is equal to

#K_c = (["C"] * ["D"])/(["A"] * ["B"]) = 1.8#

Now, if you take #x# #"M"# to be the concentration of #"A"# that is consumed by the reaction, you can say that the reaction will also consume #x# #"M"# of #"B"# and produce #x# #"M"# of #"C"# and #x# #"M"# of #"D"#.

So you can say that the equilibrium concentrations of the four chemical species will be

#["A"] = (1.00 - x) quad "M"#

#["B"] = (2.00 - x) quad "M"#

#["C"] = x quad "M"#

#["D"] = x quad "M"#

This means that the expression of the equilibrium constant will take the form

#K_c = (x * x)/((1.00-x)(2.00-x))#

#1.8 = x^2/((1.00-x)(2.00-x))#

Rearrange to quadratic equation form to get

#0.2x^2 + 5.40 * x - 3.60 = 0#

This quadratic equation will produce two solutions, one positive and one negative. Since we've taken #x# to represent concentration, you can discard the negative solution and say that

#x = 0.651#

This means that the equilibrium concentration of #"D"# will be equal to

#["D"] = color(darkgreen)(ul(color(black)("0.651 M")))#

The answer is rounded to three sig figs.