Proof Writing & Problem Solving · Easy problem

A proof that $n^2$ leaves remainder $1$ on division by $3$ whenever $3 \nmid n$ treats the case $n = 3k - 1$ by writing $n^2 = 9k^2 - 6k + 1 = 3(3k^2 - 2k) + 1$. For $n = 14$, what is $3k^2 - 2k$?