To solve the expression $f(t^2 - 2)$, we need to substitute $t^2 - 2$ into the function $f(x)$ instead of $x$. The function is given as:

$f(x) = \sqrt{2x - 3} + 4$

Now, substitute $x = t^2 - 2$ into the function:

$f(t^2 - 2) = \sqrt{2(t^2 - 2) - 3} + 4$

Let's simplify the expression inside the square root:

$2(t^2 - 2) = 2t^2 - 4$ $2t^2 - 4 - 3 = 2t^2 - 7$

Thus, we have:

$f(t^2 - 2) = \sqrt{2t^2 - 7} + 4$

So the result is:

$f(t^2 - 2) = \sqrt{2t^2 - 7} + 4$

Do you need further clarification or more details about this solution?

Here are some follow-up questions:

1. What are the domain restrictions for this function?
2. How would you graph this function?
3. Can you find the value of $f(t^2 - 2)$ for specific values of $t$?
4. How do the transformations of the square root function affect the graph?
5. What happens when $t = 1$?

Tip: Always simplify expressions inside functions before substituting to avoid mistakes in calculation.