According to Wikipedia,
Formally, an algebraic function in $n$ variables over the field $K$ is an element of the algebraic closure of the field of rational functions $K(x_1,\ldots,x_n)$.
Here's my best shot at (informally) explaining the terms in the formal definition:
A ring is a structure in which we can add and multiply.
A field is a ring in which we also can divide by any non-zero number.
When $K$ is a field, the polynomial ring in $n$ variables, denoted by the expression $K[x_1,\ldots,x_n]$, is the collection of polynomials in the variables $x_1,\ldots,x_n$ whose coefficients are in $K$.
The field of rational functions $K(x_1,\ldots,x_n)$ is defined to be $K(x_1,\ldots,x_n)=\left\{\,\frac{f}{g}\,\Bigg|\,\,\,f,g\in K[x_1,\ldots,x_n], g\neq0\right\},$ or in other words, the fractions of polynomials in $K[x_1,\ldots,x_n]$. The name "rational functions" makes sense because they are "ratios" of polynomials; in fact, the field $K(x_1,\ldots,x_n)$ stands in the same relation to the ring $K[x_1,\ldots,x_n]$ as does the field $\mathbb{Q}$ of rational numbers to the ring of integers $\mathbb{Z}$.
When $L$ is a field, the algebraic closure of $L$ is the field $\overline{L}$ that consists of all the solutions to polynomials in $L[x]$.
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For example, not every polynomial in $\mathbb{R}[x]$ (where $\mathbb{R}$ denotes the real numbers) has a solution in $\mathbb{R}$ - for example, $x^2+1=0$ has no solutions in $\mathbb{R}$. But the set of all roots of polynomials in $\mathbb{R}[x]$ forms a bigger field, containing $\mathbb{R}$ - namely $\mathbb{C}$, the complex numbers! So $\mathbb{C}=\overline{\mathbb{R}}$.
Let's consider the case of rational functions. Let $L=K(x_1,\ldots,x_n)$, and consider $L[\,t\,]$, where $t$ is a variable. The polynomial $t^2-x$ has no roots in $L$; the roots are $\sqrt{x}$ and $-\sqrt{x}$, but these don't live in $L$. They do, however, live in $\overline{L}$, which is the field of algebraic functions.