Find the value of constant $k$ that makes this function continuous The function is
$$h(x)=\begin{cases}
\dfrac{4x^2+5x-6}{x+2}  & \text{if $x\neq-2$}, \\[6pt]
3x+k &  \text{if $x= -2$}. \\
\end{cases}$$
After factoring the fraction I am left with $4x-3= -11$ when $x=-2$. If I multiply $(x+2)(4x-3)$ and then plug $-2$ into the resulting equation I am left with $-24$. If I do $4(-2)-3=3(-2)+k$ I am left with $k=-5$ which is none of the answer choices.
The answer choices to this question are
A. $k=20$
B. $k=0$
C. $k=2$
D. $k=4$
I am not sure where I go from here so any helpful tips on where I should start next would be appreciated. 

The right value is $k=-5$ as you have found.
Observing that, as $x \neq-2$, you have
$$
f(x)=\frac{4x^2+5x-6}{x+2}=\frac{(4x-3)(x+2)}{x+2}=4x-3
$$ which is equal to $-11$ as $x \to -2$. On the other hand putting $x=-2$ in $3x+k$ gives $-6+k$ then you will need that
$$
-6+k=-11
$$ that is $k=-5$.
