Truth of $x^2-2=0$
He is saying the following.
- $\sqrt{2}$ is defined as "the number such that its square is two."
- The statement "the solution $x$ to $x^2-2 = 0$ is $\sqrt{2}$" is therefore saying "the solution $x$ to $x^2-2=0$ is the number such that its square is two."
As you can see, this is a rather circular statement. This doesn't mean that irrational numbers are meaningless (indeed, $\sqrt{2}$ does exist -- see the comments and @EthanBolker's answer for more on this), but rather saying that this method of definition limits us to statements like "the number such that its square is two" or "half of the number such that its square is eight."
To elaborate on @fractal1729 's lovely correct answer:
Having defined $\sqrt{2}$ as "the number such that it's square is $2$" it's reasonable to ask whether there is such a number.
The Greeks knew that the answer is "no" if "number" means "rational number". So in order to claim the existence of $\sqrt{2}$ you must enlarge the system of rational numbers. The Greeks essentially sidestepped that question by doing geometry rather than arithmetic, using points on a line instead of numbers. Clearly there's a point on the "number line" with the right length since you can draw the diagonal of a unit square.
In later centuries mathematicians found more formal algebraic ways to enlarge the system of rational numbers to include what we now call all the "real numbers". The new ones are the irrationals. There are several ways to do this. One of the most common is to make precise the notion of an infinite decimal, along with rules for arithmetic with them. Others come with the names "Cauchy sequences" or "Dedekind cuts".
Using definite descriptor notation, we can define:
$$\sqrt{x} = (\iota y \in \mathbb{R})(y \geq 0 \wedge y^2=x)$$
In words: $\sqrt{x}$ is the unique $y \geq 0$ such that $y^2=x$.
From this, you can show that for all $y$ and all $x \geq 0$, we have: $$y^2 = x \iff y \in \pm\sqrt{x}.$$
But in some sense, this is a purely logical construct, at least insofar as we haven't explained how to approximate $\sqrt{x}$ to arbitrary precision. There's ways of doing this, but we haven't given one.