$$y=\begin{cases}\Large{\frac{\left(e^{(k+2)x}-1\right)}{5x}}&x<0\\x^2+5k-2&x\ge0\end{cases}$$
Applying the continuity theorem, I tried this $$\lim_{x\to0^-}\left(\frac{\left(e^{(k+2)x}-1\right)}{5x}\right)=\lim_{x\to 0^+}(x^2+5k-2)$$ And then seeing that the first limit was irrational I used L'Hôpital's rule and I got this $$\frac{\left((k+2)e^{(k+2)x}\right)}5$$ After that I just replaced the $x$ term with $0$ and found that $k = 1/2$.
Did I do it right?
Yes. It's correct.
$$\frac{k+2}{5}=5k-2$$
$$k+2=25k-10$$
$$12=24k$$
$$k=\frac12$$