Bất phương trình left( sqrt 3 right)2x2 + 4x ge 34x + 3?

Txđ: $D = $ $\mathbb{R}$.
$\begin{array}{l}
{\left( {\sqrt 3 } \right)^{2{x^2} + 4x}} \ge {3^{4x + 3}}\\
\Leftrightarrow {\left( {{3^{\frac{1}{2}}}} \right)^{2{x^2} + 4x}} \ge {3^{4x + 3}}\\
\Leftrightarrow {3^{{x^2} + 2x}} \ge {3^{4x + 3}}\\
\Leftrightarrow {x^2} + 2x \ge 4x + 3\\
\Leftrightarrow {x^2} - 2x - 3 \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le - 1
\end{array} \right..
\end{array}$