The expression $2(x-3)^{2}-12$ can be re-written as $a x^{2}+b x+c$ for some numbers $a, b, c$. What is the value of $10 a-b-4 c$ ?
Expanding and simplifying,

$$
2(x-3)^{2}-12=2\left(x^{2}-6 x+9\right)-12=2 x^{2}-12 x+6
$$

Thus, $a=2, b=-12$, and $c=6$.

This means that $10 a-b-4 c=10(2)-(-12)-4(6)=8$.