Probability that at least one of 2 balls taken randomly from a pile of 2 red and 3 black is red

The first solution is correct. The second solution makes the mistake of counting twice the scenario where both red balls are picked, which has a $\frac1{10}$ chance of occurring. Subtracting this from $\frac8{10}$ yields the correct answer of $\frac7{10}$.

You have double counted in your second solution. Call the red balls R1 and R2. Then you have counted both of the following:

• pick R1, then pick another ball which turns out to be R2;
• pick R2, then pick another ball which turns out to be R1.

But these choices are the same, so you should not have counted them twice. You can use your other approach, or note that this event with probability $\frac1{10}$ has been counted twice, so you should deduct it once from your answer to get the correct answer $$\frac8{10}-\frac1{10}=\frac7{10}\ .$$

For what it's worth, 7/10 (aka 14/20, reduced) is correct through another method:

I would've made this a comment since it doesn't truly answer your question, but comments can't have pretty pictures.