As, $\omega$ is complex root of unity, then
$1+\omega+\omega^2=0, \omega^3=1$
$\therefore \quad(\omega x+2)\left(\omega^2 x+2\right)-3$
$=\omega^3 x^2+2 \omega x+2 \omega^2 x+4-3$
$=x^2+2 x\left(\omega+\omega^2\right)+1 \quad\left(\because \omega^3=1\right)$
$=x^2+2 x(-1)+1=x^2-2 x+1$
$\therefore \quad \sum_{x=1}^{10}\left((\omega x+2)\left(\omega^2 x+2\right)-3\right)$
$=\sum_{x=1}^{10}\left(x^2-2 x+1\right)=\sum_{x=1}^{10} x^2-2 \sum_{x=1}^{10} x+\sum_{x=1}^{10} 1$
$=\frac{10(10+1)(2 \times 10+1)}{6}-\frac{2 \times 10(10+1)}{2}+10$
$=\frac{10 \times 11 \times 21}{6}-10 \times 11+10=385-100=285$