$x^2+2x+3≥2x^2-3x-3$
$⇔x^2+2x+3-2x^2+3x+3≥0$
$⇔-x^2+5x+6≥0$
$⇔x^2-5x-6≥0$
$⇔x^2+x-6x-6≥0$
$⇔(x+1)(x-6)≥0$
\(\Leftrightarrow\left[ \begin{array}{l}\begin{cases}x+1\geq0\\x-6\geq0\end{cases}\\\begin{cases}x+1\leq0\\x-6\leq0\end{cases}\end{array} \right.\)
\(\Leftrightarrow\left[ \begin{array}{l}\begin{cases}x\geq-1\\x\geq6\end{cases}\\\begin{cases}x\leq-1\\x\leq6\end{cases}\end{array} \right.\)
\(\Leftrightarrow\left[ \begin{array}{l}x\ge6\\x\le-1\end{array} \right.\)
Vậy với $x\ge6$ hoặc $x\le-1$ thỏa mãn.
