Thanks for the hints and solutions provided here.
Here is a rigorous proof and a summary to the hints and answers mentioned by others. Please check if its correct.
$ \mathbb{Q}[x] / (x^2 - 5) $ = {$f(x) + \langle x^2 - 5\rangle | f(x)\text{ belongs to }\mathbb{Q}[x]$}
Since $ x^2 - 5 = 0 $ (i.e. Zero Coset)
$\implies x^2 - 5 = 0 $
$\implies x^2 = 5 $
$\implies x = \pm \sqrt{5} $
So $\mathbb{Q}[x]/(x^2 - 5)$ is isomorphic to $\mathbb{Q}[\sqrt{5}]$ -- (1)
Similarly $\mathbb{Q}[x] / (x^2 + 5)$ = {$f(x) + \langle x^2 + 5\rangle | f(x) \text{ belongs to }\mathbb{Q}[x]$}
Since $x^2 + 5 = 0$ $\implies x = \pm \sqrt{5}i $
$\implies \mathbb{Q}[x]/(x^2 + 5)$ is isomorphic to $\mathbb{Q}[\sqrt{5}, i]$
Since degree of splitting field :
$[\mathbb{Q}(\sqrt{5} : \mathbb{Q}] = 2$
$[\mathbb{Q}(\sqrt{5}, i) : \mathbb{Q}] = 4$
And since the degree of the extension is not same, they are not isomorphic.
If its correct then how can I make it more rigorous?