Answer
Set $h( 2 )=7$
Work Step by Step
We see that $h(2) $ is not defined, and h is continuous everywhere else.
We check if $\displaystyle \lim_{t\rightarrow 2}h(t)=L$ exists, and if it does, we can define the continuous extension of $h$ at $t=2$ as
$H(t)=\left\{\begin{array}{ll}
h(t) , & \text{if }t\neq 2\\
L, & \text{if }t=2
\end{array}\right.$
$L= \displaystyle \lim_{t\rightarrow 2}g(x)= \displaystyle \lim_{x\rightarrow 3}\frac{t^{2}+3t-10}{t-2}\quad $...factor the trinomial,
$= \displaystyle \lim_{t\rightarrow 2}\frac{(t+5)(t-2)}{(t-2)}\quad $...cancel common term
$= \displaystyle \lim_{t\rightarrow 2}(\mathrm{t}+5)$
... evaluate directly,
$L=7$
The limit exits at $t=2,$ so, if we redefine $h$,
$h(t)=\left\{\begin{array}{ll}
\dfrac{t^{2}+3t-10}{t-2} , & \text{if }t\neq 2\\
7, & \text{if }t=2
\end{array}\right.$
it becomes continuous.