Voltage divider bias (BJT) leads to a huge voltage drop across the collector resistor. Why?
If you getting such a strange result that's mean that your BJT is in the saturation region. 
In your circuit, the collector current cannot be larger than: 
$$I_{Cmax} = \frac{V_{CC}}{R_C+R_E} \approx 3.2mA $$ and you get \$8.2mA\$ hence your BJT is in the saturation region.

So, to be able to solve it without using the advanced math you are forced to assume some \$V_{CEsat}\$ value. Typically in hand calculations, we are assuming that the saturation voltage is equal to \$V_{CEsat} \approx 0.2V\$ 

Additional we use this equation \$I_E=I_B+I_C\$ which is always true is saturation and in the active region.

$$I_E = \frac{V_E}{R_E}$$ 

$$I_C = \frac{V_{CC} - (V_{CEsat}+V_E)}{R_C} $$

$$I_B= \frac{V_{BB}-(V_{BE}+ V_E)}{R_{th}}$$

So finally we have this equation with only one unknown \$V_E\$ 

$$\frac{V_E}{R_E}=\frac{V_{BB}-(V_{BE}+ V_E)}{R_{th}}+\frac{V_{CC} - (V_{CEsat}+V_E)}{R_C}$$

And the solution is 

$$V_E = \left(\frac{V_{BB} - V_{BE}}{R_{th}} +\frac{V_{CC} -V_{CEsat}}{R_C}\right)\cdot R_E||R_{th}||R_C \approx 4.49V$$

And the currents are: 

$$I_E \approx 4.49 \textrm{m}A$$

$$I_B \approx 1.79\textrm{m}A $$

$$I_C \approx 2.7\textrm{m}A $$

Or you could have written this two equations and solve for \$I_B\$ and \$I_C\$

$$V_{BB} - I_B R_B - V_{BE} - (I_B+I_C) R_E = 0$$

$$V_{CC} - I_C R_C - V_{CEsat} - (I_B+I_C) R_E = 0$$