#### Answer

See below:

#### Work Step by Step

Consider the provided inequality
$3{{x}^{2}}+16x+5<0$
The boundary points of the given inequality will be calculated by equating $f\left( x \right)$ to 0 as $3{{x}^{2}}+16x+5=0$
Thus,
$\begin{align}
& 3{{x}^{2}}+x+15x+5=0 \\
& x\left( 3x+1 \right)+5\left( 3x+1 \right)=0 \\
& \left( x+5 \right)\left( 3x+1 \right)=0
\end{align}$
Hence,
$x=-5,x=-\frac{1}{3}$
These values of x are the boundary points, so locate these point on the number line.
From the above number line, the boundary points divide the number line into three parts as,
$\left( -\infty ,-5 \right)\text{,}\left( -5,-\frac{1}{3} \right)\text{ and }\left( -\frac{1}{3},\infty \right)$.
Now, one test value within each interval is chosen and f is evaluated at that number.
As can be observed, for both intervals $\left( -\infty ,-5 \right)\text{ and }\left( -\frac{1}{3},\infty \right)$ , the function is positive.
And, for interval $\left( -5,-\frac{1}{3} \right)$ , the function is negative.
Also, the points $-5\text{ and }-\frac{1}{3}$ are not included in the solution because the function is not equal to 0.
Hence, the required interval is $\left( -5,-\frac{1}{3} \right)$.