According to Kirchhoff's current law the algebraic sum of the currents flowing into and out of a node is zero, charges cannot accumulate and cannot be created in a node. As much current flows in, as much will flow out.
$ I_1 - I_2 + I_3 = 0 $
In a closed loop, the algebraic sum of the voltages is zero. To use this law, take an arbitrary direction around the loop and calculate the voltage of each element accordingly. The voltage of resistors is the product of the resistance and the current, according to Ohm's law. The sign is determined by the relationship between the assumed current direction and the arbitrary direction. Similarly the signs of voltage sources depend on the arbitrary direction, as well. If directions are opposite, the sign becomes negative.
$ - V_1 + I_1 ⋅ (R_1 + R_2) - I_2 ⋅ (R_3 + R_4) + I_3 ⋅ R_5 = 0 $
By applying both Kirchhoff's laws to the following two-loop circuit, its currents can be calculated, since the three currents of the three branches form a system of equations that can be solved with any method.
$ V_1 = 7 \, [V] $
$ R_3 = 10 \, [Ω] $
$ V_2 = 9 \, [V] $
$ R_4 = 12 \, [Ω] $
$ R_1 = 6 \, [Ω] $
$ R_5 = 14 \, [Ω] $
$ R_2 = 8 \, [Ω] $
$ R_6 = 16 \, [Ω] $
$ -V_1 + I_1 ⋅ (R_1 + R_2) + I_2 ⋅ (R_3 + R_4)= 0 $
$ V_2 - I_2 ⋅ (R_3 + R_4) - I_3 ⋅ (R_5 + R_6) = 0 $
$ 14 ⋅ I_1 + 22 ⋅ I_2 = 7 $
$ - 22 ⋅ I_2 - 30 ⋅ I_3 = -9 $
$ I_1 - I_2 + I_3 = 0 $
$ \begin{bmatrix} 14 & \hfill 22 & \hfill 0\\ \hfill 0 & -22 & -30 \\ \hfill 1 & \hfill -1 & \hfill 1 \end{bmatrix} ⋅ \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} \hfill 7 \\ -9 \\ \hfill 0 \end{bmatrix} $
$ I_1 = 0.1196 \, [A] = 119.6 \, [mA] $
$ I_2 = 0.2421 \, [A] = 242.1 \, [mA] $
$ I_3 = 0.1225 \, [A] = 122.5 \, [mA] $