[Coastie17]

Concentration (M)   Fe3+         SCN-       FeSCN2+
Initial                       4.0 x 10^-5    0.50           0
Completion                     0         0.50      4.0 x 10^-5
∆, reverse rxn            +x          +x           -x
equilibrium                     x      0.50 + x      4.0 x 10^-5 - x

The initial concentrations of Fe3+ and SCN- are correct and the completion concentration of FeSCN2+ is also correct. I am unsure of the completion value for SCN-, though because the initial concentration of Fe3+ and the completion concentration of FeSCN2+ are the same, I am assuming that the SCN- concentration = 4.0 x 10^-5 - 0.50 = 0.50.
So, the question is asking to find the equilibrium concentrations for the reverse reaction using the K value of 1.6 x 10^2. Here are my calculations for the answers in the table.

(x)(x +0.50)/ (4.0 x 10^-5 - x) = 1.6 x 10^2
x^2 + 0.50x = 6.4 x 10^-3 - (1.6 x 10^2)x
x^2 + (1.6 x 10^2)x - 6.4 x 10^-3 = 0

Using  the quadratic formula:
-1.6 x 10^2 ± √(1.6 x 10^2)^2 - 4(1)(-6.4 x 10^-3) = x
-1.6 x 10^2 ±√2.6 x 10^4 - (-2.6 x 10^-2) = x
-1.6 x 10^2 ±√2.6 x 10^4 = x
-1.6  x 10^2 ± 1.6 x 10^2 = x
0 = x

[Fe3+] = x = 0
[SCN-] = 0.50 + x = 0.50 + 0 = 0.50
[FeSCN2+] = 4.0 x 10^-5 - x = 4.0 x 10^-5 - 0 = 4.0 x 10^-5

So, I guess my question is, does this look correct or do you see anything I did wrong? Thanks!

[DevaDevil]

Here are my calculations for the answers in the table.

(x)(x +0.50)/ (4.0 x 10^-5 - x) = 1.6 x 10^2
x^2 + 0.50x = 6.4 x 10^-3 - (1.6 x 10^2)x
x^2 + (1.6 x 10^2)x - 6.4 x 10^-3 = 0

do not forget to bring over the whole number of 0.5+1.6 x 10²!
Plus: do not lose the extra digits until you check significant numbers at the end.

Using  the quadratic formula:
-1.6 x 10^2  √(1.6 x 10^2)^2 - 4(1)(-6.4 x 10^-3) = x

abc formula:  x = ( -b ± √{b²-4ac} ) / 2a


if you correct these 2 things, you will get a different x value