Is this an extraneous solution? Does $$\sqrt{2x} = \sqrt{5x+3}$$ have an extraneous solution which is $-1$ or is the solution $-1$ ? As I solved, both sides become $\sqrt{-2}$ but negative numbers can't be in the square root.

Generally, we would say this is an extraneous solution. Indeed, if we look at the domains of our functions beforehand, you ought've noticed that

$$\sqrt{2x}\implies x\ge0$$

$$\sqrt{5x+3}\implies x\ge-\frac35$$

Combined, we can only have $x\ge0$, so $x=-1$ is not a solution here.