How do you prove glycine is 90 percent protonated at pH 8.6 using the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation states that the pH is equal to the pKa plus the ratio of the deprotonated form to the protonated form of an ionizable compound. (pH=pKa+log([A^-]/[HA])). Glycine has two ionizable side groups; the amino group (pKa ~9.6) and the carboxylic acid group (pKa ~2.2).

At pH = 8.6, the carboxlyic acid group will be mostly deprotonated (99.999%) because
pH=pKa+log([A^-]/[HA])
let x=the concentration of A^- (deprotonated carboxylic acid). [HA] then must equal 1-x because the concentration of A^- and HA together must be 1.
8.6=2.2+log(x/(1-x))
6.4=log(x/(1-x))
10^6.4=(x/(1-x))
(1-x)*10^6.4=x
10^6.4-10^6.4x=x
10^6.4=x+10^6.4x=(10^6.4+1)x
10^6.4/(10^6.4+1)=x=0.99999=99.999%

Using the same mathematical approach, we can see for the amino group
8.6=9.6+log(x/(1-x))
-1=log(x/(1-x))
10^-1=(x/(1-x))
(1-x)*10^-1=x
10^-1-10^-1x=x
10^-1=x+10^-1x=(10^-1+1)x
10^-1/(10^-1+1)=x=0.11=11.0%

So the amino group is in fact 11% deprotonated at this pH (89% protonated).