Okay, so I need to find the point that the curve passes through when t is 2. The curve is given by the parametric equations (x, y) = (t³ + 7, -3t² - 6t - 5). Alright, let me think about how to approach this.

First, since it's a parametric curve, each value of t gives a specific point (x, y). To find the point at t=2, I just need to substitute t=2 into both the x and y equations and compute the results. That seems straightforward, but let me make sure I don't make any calculation errors.

Starting with the x-coordinate: x = t³ + 7. If t=2, then x = 2³ + 7. Calculating 2³ is 8, so 8 + 7 equals 15. So the x-coordinate should be 15. Let me double-check that. 2 cubed is indeed 8, adding 7 gives 15. That seems right.

Now for the y-coordinate: y = -3t² - 6t - 5. Substituting t=2 here. Let's compute each term step by step. First, t² when t=2 is 4. Then, -3 times 4 is -12. Next, -6 times 2 is -12. Adding those two terms: -12 + (-12) = -24. Then, subtract 5: -24 - 5 = -29. Wait, is that correct? Let me verify again.

Breaking it down:

-3t²: when t=2, that's -3*(2)² = -3*4 = -12.

-6t: when t=2, that's -6*2 = -12.

Then the constant term is -5.

Adding them all together: -12 (from -3t²) + (-12) (from -6t) + (-5) = -12 -12 -5. Let's add those: -12 -12 is -24, then -24 -5 is indeed -29. So the y-coordinate is -29.

Therefore, putting it all together, the point on the curve when t=2 is (15, -29). Let me just check once more to be sure.

For x: 2³ is 8, plus 7 is 15. Correct.

For y: -3*(2)² is -12, minus 6*(2) is -12, minus 5. So total is -12 -12 -5 = -29. Yep, that's right.

I think that's all. No complicated steps here, just substitution and arithmetic. The calculations seem straightforward and I double-checked each part. So the final answer should be (15, -29).
Therefore, the curve passes through the point $\boxed{(15, -29)}$ at $t = 2$.
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