Solve for Y 1

$y^2-4=x$ (Add $4$ to both sides of the equation.)

$y^2-4+4=x+4$ (Simplify)

$y^2=x+4$ (Extract the square root of both sides of the equation to isolate $y$ .)

$\sqrt{y^2}=\pm \sqrt{x+4}$

$y= \pm \sqrt{x+4}$

The positive solution is,

$\boxed{y=\sqrt{x+4}}$

Here are the steps to solve the equation for $y$ .

$y^2 - 4 = x$ $y^2 - 4 + 4 = x + 4$ $y^2 = x + 4$ $\sqrt{y^2} = \sqrt{x +4}$ $y = \pm \sqrt{x +4}$ So the positive solution is $y = \sqrt{x +4}$

Isolating $y$ gives: $\begin{aligned} y^2 - 4 &= x \\ y^2 &= x + 4 \\ y &= \pm \sqrt{x + 4} \end{aligned}$ The positive solution is $\boxed{\sqrt{x + 4}}$ .

Therefore, the answer is $\sqrt{x + 4}$ .