When $x$ tends to $3$ is $x+3$ always less than $6$ or not?

In a question, it was asked to prove that as limit $x$ goes to $3$, $x^{2}$ tends to $9.$

Now, on solving it by the $\epsilon$-$\delta$ method, when I simplified $x^{2} - 9$ to $\left(x-3\right)\left(x+3\right)$ and put $x+3$ to be $6$ because $x+3$ will always be less than $6$ (to get a suitable $\delta$) and by this, my $\delta$ came out to be $\epsilon/6.$

But in the actual solution they used that $x+3$ will be less than $7$ as $x$ tends to $3$. But isn't choosing $6$ more accurate or am I missing something?