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

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