[tex]-3x^2+12y^2=84[/tex]
[tex]12y^2=3x^2+84[/tex]
[tex]y^2=\dfrac{x^2}4+7[/tex]
[tex]y=\pm\sqrt{\dfrac{x^2}4+7}[/tex]
For either square root to exist, you require that [tex]\dfrac{x^2}4+7\ge0[/tex]. This is true for all [tex]x[/tex], since [tex]\dfrac{x^2}4[/tex] is always non-negative. This means the domain of [tex]y[/tex] as a function of [tex]x[/tex] is all real numbers, or [tex]x\in\mathbb R[/tex] or [tex](-\infty,infty)[/tex].
Now, because [tex]\dfrac{x^2}4+7[/tex] is non-negative, and the smallest value it can take on is 7, it follows that the minimum value for the positive square root must be [tex]\sqrt7[/tex], while the maximum value of the negative root must be [tex]-\sqrt7[/tex]. This means the range is [tex]y\in\mathbb R\setminus(-\sqrt7,\sqrt7)[/tex], or [tex]|y|\ge\sqrt7[/tex], or [tex](-\infty,-\sqrt7]\cup[\sqrt7,\infty)[/tex].
