Answer by periblepsis for Find equivalent resistance of a resistor circuit

There isn't just one process. There's many different approaches. Some more general than others.

But in this case I'd like to call your attention to the voltage dividers present in your circuit.

Note that \$R_3\$ and \$R_5\$ form one voltage divider and that \$R_1\$ and \$R_4\$ form another voltage divider. So Thevenize \$R_3\$ and \$R_5\$ and then Thevenize \$R_1\$ and \$R_4\$. You now have:

^{simulate this circuit– Schematic created using CircuitLab}

This tells you the two node voltages.

From this, you only need to sum up the two currents, the two present in \$R_1\$ and \$R_3\$ since those are the only current paths that the voltage source can take, and divide that sum into the source voltage, \$V=20\:\text{V}\$:

$$\begin{align*}R_\text{EQ}&=\frac{V}{\frac{V-\left(V_{\text{TH}_1}-I_\text{TH}\,\cdot\,R_{\text{TH}_1}\right)}{R_3}+\frac{V-\left(V_{\text{TH}_2}+I_\text{TH}\,\cdot\,R_{\text{TH}_2}\right)}{R_1}}\\\\&=\frac{20\:\text{V}}{\frac{20\:\text{V}-12\:\text{V}}{10\:\Omega}+\frac{20\:\text{V}-8\:\text{V}}{30\:\Omega}}=\frac{20\:\text{V}}{800\:\text{mA}+400\:\text{mA}}\end{align*}$$

And you are done.