Find H3O+ of weak acid salt solution which compose of solute that can gain and give protons

Find $[\ce{H3O+}]$ of 0.100 M $\ce{NaHCO3}$ ($K\_{\ce{H2CO3},1}=4.45×10^{-7}$, $K\_{\ce{H2CO3},2}=4.7×10^{-11}$)


My work is: $$\ce{HCO3- + H2O<=>H2CO3 + OH-}$$


Initial: $[\ce{HCO3-}]$ = 0.100 M, $[\ce{H2CO3}]$ = 0, $[\ce{OH-}]$ = 0  

Change: $[\ce{HCO3-}]$ = -X, $[\ce{H2CO3}]$ = +X, $[\ce{OH-}]$ = +X  

Equilibrium: $[\ce{HCO3-}]$ = 0.100 - X, $[\ce{H2CO3}]$ = X, $[\ce{OH-}]$= X 


$K\_{b2}=\frac{Kw}{K\_{a1}} = 2.25×10^{-8}$  

$K\_{b2} < 10^{-3}: 0.100 - x = x$  

$K\_{b2} = \frac{x^2}{0.100}$  

$X = [\ce{OH-}] = 4.47×10^{-5}$ so $[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]}$, $[\ce{H+}] = 2.11×10^{-10}$


And the second equation is 
 $$\ce{HCO3- + H2O<=>CO2 + H3O+}$$
For this equation $K\_{a2}=4.7×10^{-11}$  

So when I compared $K$ value between both of these equation, I will get $K\_{b2} >> K\_{a2}$ (more than $10^3$ times). So I think increase $[\ce{H3O+}]$ which occur or $[\ce{CO2}]$ can be ignored. So my answer $[\ce{H3O+}]$ is equal to $2.11×10^{-10}$ M.


**But** the solution in my book is:  

From the formula:  

$[\ce{H3O+}]= \sqrt{\dfrac{K\_1 K\_2 × [\ce{NaHCO3}] + k\_1 k\_w}{[\ce{NaHCO3}] +K\_1}}$


$[\ce{H3O+}]=4.6×10^{-9}$ (solution answer)


Do I approximately second equation false or $K$ value is false or another thing?


