Finding the (Q) point for V(Collector-Emitter) and Collector Current?
The most important thing to recognize in this circuit, is that the current through R1 and R2 is not equal and that everything can be solved algebraically. Let's just write down everything we know and see where we land. In this notation, \$I_x\$ means the current through x


\$(I_{R_2}+I_B)R_1 + I_{R_2}R_2 = 20 V\$
\$(I_B+I_C)R_E + V_{CE} + I_CR_C = 20 V\$
\$I_C = \beta * I_B\$
\$(I_B + I_C)Re + V_{BE} + (I_{R_2} + I_B)R1 = 20 V\$


This system of equations has too many unknowns: \$I_{R_2}, I_B, I_C, V_{BE}\$ and \$V_{CE}\$. We need to either find a fourth independent equation or guess one of them. I assume that you are being expected to assume that \$V_{BE}\$ is constant, as per your statement that it is 0.7V. This allows you to solve this system of equations:


\$I_B = (20 - 0.7 - {R_1* 0.7\over R_2})/(R_1 + R_E*(1+\beta) + {Re*R_1*(1 + \beta)\over R_2}) = 28.75904 µA\$
\$I_C = 50*I_B = 1.43795245 mA\$
\$I_E = I_B + I_E = 1.46671149 mA\$
\$V_{CE} = 20 - I_ER_E - I_CR_C = 20 - 1.613382639 - 8.91530519 = 9.471312171 V \$