This is the working to find the gcd using the Euclidean algorithm $$\begin{align}1820 &= 7(231) + 203 & (a) \\ 231 &= 1(203) + 28 &(b)\\ 203 &= 7(28) + 7 &(c)\\ 28 &= 4(7)+0 &(d)\\ \end{align}$$
The last non-zero remainder is 7, so gcd(1820, 231) = 7.
And then to find linear combination m and n:
$$\begin{align} 7 &= 203 − 7(28) &\mathrm{from}\ (c) \\ &= 203 − 7(231 − 203) &\mathrm{from}\ (b) \\ &= (−7)(231) + 8(203) \\ &= (−7)(231) + 8(1820 − 7(231)) &\mathrm{from}\ (a)\\ &= 8(1820) + (−63)(231)\\ \end{align} $$
However I'm confused at the linear combination line where it has = (−7)(231) + 8(203). Where did the 8 come from in this line? There was no 8 in the previous line.
The extended Euclidean algorithm is more easily done this way, in tabular form:
$$\begin{matrix} 1820 & 1 & 0 & -\\ 231 & 0 & 1 & 7 \\ 203 & 1 & -7 & 1 \\ 28 & -1 & 8 & 7 \\ 7 & 8 & -63 & 4 \\ 0 & (-33) & (260) & - \\ \end{matrix}$$
where the first column are the remainders you computed, the final column the quotients, and the two in the middle are the coefficients of $1820$ and $231$ respectively.
To compute 203 we do $1820 - 7 \times 231$ so the coefficients are found using the same relation on the previous rows: a row with $1$ and $0$ (for $1820$) minus 7 times the row with $0$ and $1$, giving $1$ (for $1 - 7 \times 0$) and $-7$ (for $0 - 7 \times 1$) hence the third row. 203 goes once into 231 so we have $0\ 1$ minus $1$ times $0 \ -7$ giving $-1 \ 8$.
We go on that way (e.g. $-63 = -7 - 7 \times 8$ etc.) until we reach $0$ on the left (no need to do those coefficients; I put them between braces) and then the row before it has the gcd and its coefficient combination:
$$7 = 8 \times 1820 + (-63) \times 231$$
This avoids the error-prone back-substitutions and is easy to program.