Literature value of Ka = 7.20 x 10^-4 = [H3O^+][F^-]/[HF] =
(x)(0.15 + x)/(0.15- x) = 7.20 x 10^-4
0.15(x) + x^2 = 7.20 x 10^-4(0.15 - x) = 1.66 x 10^-4 - 7.20 x 10^-4(x)
This can be solved as a quadratic equation. See the 2nd link.
x^2 + 0.15(x) - 1.66 x 10^-4 = 0
x = 1.10 x 10^-3 M = [H3O^+]