$\varepsilon$-$\delta$ definition of a limit [Calculus] $$\lim_{x\to 2} x^2 - 8x + 8= -4 $$
Now to express $ f(x) - l $ it as $x - x_0$ , we write $ x = (x - x_0) + x_0$.
So \begin{align}
f(x) - l &= x^2 - 8x + 8 + 4\\
&= (x - 2 + 2)^2 - 8(x - 2 + 2)+ 12\\
&= (x - 2)^2 +4(x-2) + 4 - 8(x - 2) - 16+ 12 \\
&= (x - 2)^2 - 4(x-2)\\
&= |f(x) - l|\\
&\le |x - 2|^2 -4|x-2|
\end{align}
We choose $\delta$ such that $\delta^2 -4\delta \le \varepsilon$.
For $\delta \le 1$,  $$\delta -4\delta \le \varepsilon$$
$$\delta \le -\frac\varepsilon3$$
Is my $\delta$ correct ? Could someone guide me through this? Only then I guess I can verify this limit by $\varepsilon$-$\delta$ definition. Thanks.

No, it is not correct. If you choose $\delta\leqslant-\frac\varepsilon3$, then $\delta<0$. And it is part of the $\varepsilon-\delta$ definition of limite that $\delta>0$.
Note that the inequality $\bigl|f(x)-l\bigr|\leqslant|x-2|^2-4|x-2|$ is false, since it leads to situations in which $\bigl|f(x)-l\bigr|<0$.
