Why is my solution incorrect for solving these quadratic equations?

You have to pay attention to your domain. In the first equation you get a positive and a negative value for $y$, while $\sqrt{x}$, which is your substitution, can only be positive.

You are doing your algebra correctly, the only problem is when you have a positive and a negative value for a square root you have to ignore the negative one.

I checked the answer $$ x= \frac {58-10 \sqrt {33}}{4}$$ for your first problem and it does work nicely.

Based on comments some of your issues seemed to be in checking answers and others likely an issue in arithmetic (even the greatest sometimes think $1+1=3$ every now and then, but I can’t speak for them). Here’s a tip I have for checking answers to quadratics.

Generally you have something of the form:

$$Ax^2 + Bx + C = 0$$

You only need to know it is zero. So what I did was this. I didn’t multiply everything. I plugged enough in to verify. So say D = x. I might do the following:

$$DAx + Bx + \frac CD x = 0$$

Now you just take the sum $DA + B + \frac CD$ which can be done mentally with most high school practice problems.

In the case of $(x-7)^2 - 13(x-7) + 36 = 0$ this makes it much easier to verify $11$ and $16$ as answers.

As for general quadratics your style and technique is good. I don’t see any flaws in your thinking aside from being careful about the back substitutions. Definitely watch for dropping negatives and adding them in. Doing your own new problem might introduce needs to carefully think about the Complex Plane if you aren’t careful. However picking the right substitutions is an art and one I cannot really describe with a process or answer. Most people at your level do not substitutions and so your definitely thinking outside the box. This is a good thing. I did not even know if substitutions until Calculus and it made it very hard to grasp that concept (of just creating a new variable on the fly) so kudos to you.