Answer
$$\frac{-3}{4} $$
Work Step by Step
Given $$ \lim _{x \rightarrow -\infty}(\sqrt{4 x^{2}+3x}+2 x)$$
Then
\begin{aligned}
\lim _{x \rightarrow-\infty}(\sqrt{4 x^{2}+3x}+2 x) &=\lim _{x \rightarrow -\infty} \frac{(\sqrt{4 x^{2}+3x}+2 x)(\sqrt{4 x^{2}+3x}-2 x)}{ \sqrt{4 x^{2}+3x}-2 x}\\
&=\lim _{x \rightarrow -\infty} \frac{(\sqrt{4 x^{2}+3x})^2-(2 x)^2 }{ \sqrt{4 x^{2}+3x}-2 x} \\
&=\lim _{x \rightarrow -\infty} \frac{4 x^{2}+3x-4 x^2 }{ \sqrt{4 x^{2}+3x}-2 x}\\
&=\lim _{x \rightarrow- \infty} \frac{3x}{ \sqrt{4 x^{2}+3x}-2 x} \cdot \frac{1 / x}{1 / x} \\
&=\lim _{x \rightarrow -\infty} \frac{3x / x}{\sqrt{4 x^{2} / x^{2}+3x / x^{2}}-2}\\
&=\lim _{x \rightarrow- \infty} \frac{3}{\sqrt{4+3 / x}-2}\\
&=\frac{3}{-2-2}=\frac{-3}{4}
\end{aligned}