Performing DC analysis by open circuiting all capacitors

The voltage across resistor R2 can be given by the expression

The current through R1 is

$I_1=\dfrac{V_{CC}-V_B}{R_1}$

Similarly, current through R2 is

$I_2=\dfrac{V_B-V_{EE}}{R_2}$

Since there is no path for these currents to take another loop, these currents have got to be the same which brings us to the foll. equation

$\dfrac{V_{CC}-V_B}{R_1}=\dfrac{V_B-V_{EE}}{R_2}$

$\dfrac{V_{CC}-V_B}{R_1}-\dfrac{V_B-V_{EE}}{R_2}=0$

$\dfrac{V_{CC}R_2-V_BR_2-V_BR_1+V_{EE}R_1}{R_1R_2}=0$

${V_{CC}R_2-V_BR_2-V_BR_1+V_{EE}R_1}=0$

$V_{CC}R_2+V_{EE}R_1=V_B(R_1+R_2)$

$\therefore V_B=\dfrac{V_{CC}R_2+V_{EE}R_1}{R_1+R_2}$

NOTE: Using the above equation, we can easily get VB but just in case you want a easier equation, we will simplify the obtained equation

$V_B=\dfrac{V_{CC}R_2+V_{EE}R_2-V_{EE}R_2+V_{EE}R_1}{R_1+R_2}$

$V_B=\dfrac{V_{EE}(R_1+R_2)}{R_1+R_2}+\dfrac{R_2(V_{CC}-V_{EE})}{R_1+R_2}$

$\therefore V_B=V_{EE}+(V_{CC}-V_{EE})\times\dfrac{R_2}{R_1+R_2}$

$\therefore V_B=-5+\big(5-(-5)\big)\times\dfrac{2\times10^3}{12\times10^3+2\times10^3}$

$\therefore V_B=-3.571\ V$

$R_B=R_1||R_2$

$R_B=\dfrac{12\times10^3\times2\times10^3}{12\times10^3+2\times10^3}$

$\therefore R_B=1.714\ K \Omega$

Applying KVL to Base - Emitter Loop

$V_{B}-I_BR_B-V_{BE}-I_ER_E+V_{EE}=0$

But, $I_E=(1+\beta)I_B$

$V_B-I_BR_B-V_{BE}-(1+\beta)I_BR_E+V_{EE}=0$

$V_B-V_{BE}+V_{EE}=I_B\big[R_B+(1+\beta)R_E\big]$

$I_B=\dfrac{V_B-V_{BE}+V_{EE}}{R_B+(1+\beta)R_E}$

$I_B=\dfrac{-3.571-0.7+6}{1.71\times10^3+(1+100)500}$

$I_B=33.135\ \mu A$

We know that, $I_C=\beta I_B$

$I_C=100\times34.237\times10^{-6}$

$\therefore \underline{\underline{I_C=3.135\ mA}}$

Applying KVL to the Collector - Emitter Loop

$V_{CC}-I_CR_C-V_{CE}-I_ER_E+V_{EE}=0$

But $I_C\approx I_E$

$\therefore V_{CC}-I_C\big[R_C+R_E\big]-V_{CE}+V_{EE}=0$

$V_{CE}=V_{CC}-I_C\big[R_C+R_E\big]+V_{EE}$

$\therefore V_{CE}=5-3.135\times10^{-3}\big[5\times10^3+500\big]+5$

$\therefore \underline{\underline{V_{CE}=-8.288\ V}}$