The problem: find the vertical asymptotes of the function $f(x) = \dfrac{\sqrt{16x^2+3x+6}-5}{x-1}$.
From the formula it should be clear that if there is an asymptote, it would occur at $x=1$ (this is because that's the value of $x$ that makes the denominator zero).
So it is asking you to consider both $\displaystyle\lim_{x \rightarrow 1^+} \dfrac{\sqrt{16x^2+3x+6}-5}{x-1}$ and $\displaystyle\lim_{x \rightarrow 1^-} \dfrac{\sqrt{16x^2+3x+6}-5}{x-1}$
Let's look at the first one. The "standard trick" for a problem with a radical in it like this is the "multiply by convenient form of $1$" trick with the conjugate, so you multiply the function by $$\dfrac{\sqrt{16x^2+3x+6}+5}{\sqrt{16x^2+3x+6}+5}$$ and it will result in
$$(\dagger) \hspace{35pt} \begin{array}{ll}
\displaystyle\lim_{x \rightarrow 1^+} \dfrac{\sqrt{16x^2+3x+6}-5}{x-1} &= \displaystyle\lim_{x \rightarrow 1^+} \dfrac{\sqrt{16x^2+3x+6}-5}{x-1}\dfrac{\sqrt{16x^2+3x+6}+5}{\sqrt{16x^2+3x+6}+5} \\
&= \displaystyle\lim_{x \rightarrow 1^+} \dfrac{16x^2+3x-19}{(x-1)[\sqrt{16x^2+3x+6}+5]}
\end{array}$$
at this point you usually would factor the numerator to get something that would cancel with the $(x-1)$ in the denominator. It does not seem clear that such a factoring exists, but there is a way to tell for sure: suppose that
$$16x^2+3x-19=(x-1)(cx-b)$$
for some unknown constants $c$ and $b$. From this we see that
$$\dfrac{16x^2+3x-19}{x-1} = cx-b$$
and so we will try to figure this out by long division of $\dfrac{16x^2+3x+19}{x-1}$ (Do you remember how that is done? It is a college algebra topic.) and you will see that
$$\dfrac{16x^2+3x-19}{x-1}=16x+19.$$
Therefore you can conclude
$$16x^2+3x-19=(16x+19)(x-1)$$
and therefore with our limit marked above by $(\dagger)$, we see
$$\begin{array}{ll}
\displaystyle\lim_{x \rightarrow 1^+} \dfrac{16x^2+3x-19}{(x-1)[\sqrt{16x^2+3x+6}+5]} &= \displaystyle\lim_{x \rightarrow 1^+} \dfrac{(x-1)(16x+19)}{(x-1)[\sqrt{16x^2+3x+6}+5]} \\
&= \displaystyle\lim_{x \rightarrow 1^+} \dfrac{16x+19}{\sqrt{16x^2+3x+6}+5} \\
&= \dfrac{7}{2}.
\end{array}$$
Can you compute this limit as $x \rightarrow 1^-$?
You will see that it also equals $\dfrac{7}{2}$. From this information you can conclude that no vertical asymptote exists at $x=1$, because neither of these limits diverged to infinity.
