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ratio of partial sums of the same sequence

My kid was given this question:
> In a geometric sequence, the proportion of (the sum of the first $12$ terms) to (the sum of the first $8$ terms) is $\frac{819}{51}$. Find the common ratio of the sequence.

The only formula thus far covered for the partial sum of a geometric sequence $(a_1q_i)_{i\ge0}$ is$$\frac{a_1(q^n-1)}{q-1}.$$ How does one solve this?

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The solution I found is as follows, but it seems too roundabout for the context, so I wonder what more direct way there is:

The given information means$$\frac{q^{12}-1}{q^8-1}=\frac{819}{51}.$$ Let $u=q^4$. Then$$\frac{819}{51}=\frac{u^3-1}{u^2-1}=\frac{u^2+u+1}{u+1}$$ $$\frac{819-51}{51}=\frac{u^2}{u+1}$$ $$51u^2-768u-768=0$$so $u\in\lbrace-16/17,16\rbrace$ and $q=\pm2$.