You are looking to solve the equation:

$\sqrt{y} - \frac{1}{\sqrt{y}} = 2$

We can try solving this step by step. Let's set:

$x = \sqrt{y}$

This makes the equation:

$x - \frac{1}{x} = 2$

Now, multiply both sides of the equation by $x$ to eliminate the fraction:

$x^2 - 1 = 2x$

Rearrange the terms to form a quadratic equation:

$x^2 - 2x - 1 = 0$

Now solve this quadratic equation using the quadratic formula:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$

$x = \frac{2 \pm \sqrt{4 + 4}}{2}$

$x = \frac{2 \pm \sqrt{8}}{2}$

$x = \frac{2 \pm 2\sqrt{2}}{2}$

$x = 1 \pm \sqrt{2}$

So, $x = 1 + \sqrt{2}$ or $x = 1 - \sqrt{2}$.

Since $x = \sqrt{y}$, and $x$ must be positive, we discard $x = 1 - \sqrt{2}$ (as it's negative). Hence, $x = 1 + \sqrt{2}$.

Now, solve for $y$:

$\sqrt{y} = 1 + \sqrt{2}$

Square both sides:

$y = (1 + \sqrt{2})^2$

$y = 1^2 + 2 \cdot 1 \cdot \sqrt{2} + (\sqrt{2})^2$

$y = 1 + 2\sqrt{2} + 2$

$y = 3 + 2\sqrt{2}$

Thus, the value of $y$ is:

$y = 3 + 2\sqrt{2}$

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Would you like further details or have any other questions?

Here are some related questions to explore:

1. How do we apply the quadratic formula in different contexts?
2. What are the properties of square roots in algebraic equations?
3. How does squaring affect the sign of a number?
4. What other methods can we use to solve non-linear equations?
5. How does multiplying by the variable simplify fractional equations?

Tip: When simplifying quadratic equations, check for factorization before applying the quadratic formula—it might save you time!