Proving (or disproving) that two sets are equal

Question

I'm stuck proving (or disproving) the equality of two sets $M_1=M_2$.

$M_1=\{k\in\mathbb{N}\ |\ \exists x(x\in\mathbb{N}\ \wedge\ k^2+4k+4=2x+4)\}$

$M_2=\{l\in\mathbb{N}\ |\ \forall y(y\in\mathbb{N}\ \rightarrow y\leq l)\}$

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Here are my steps so far:

Solving for $k$ I get

$M_1=\{k\in\mathbb{N}\ |\ \exists x(x\in\mathbb{N}\ \land\ (k=-2+\sqrt{4+2x}\ \lor\ k=-2-\sqrt{4+2x}))\}$

then I transform $M_2$, such that

$M_2=\{l\in\mathbb{N}\ |\ \forall y(\lnot (y\in\mathbb{N})\ \lor (y\leq l))\}$

Now, I don't know what to do. What is the next step?

I know that, to prove the equality of two sets, I must proof that the first set is contained in the second set and vice versa. But how do I do that given the sets I have?

Answer 1

$2^2 + 4\times 2 + 4 = 16 = 2\times 6 + 4$ therefore $2\in M_1$.

Assume $l\in \Bbb N$. Then $l+1 \gt l$, thus $l \notin M_2$, therefore $M_2 = \emptyset$.

Thus $M_1 \ne M_2$.