In simple English, what does it mean to be transcendental? From Wikipedia


A transcendental number is a real or complex number that is not algebraic
A transcendental function is an analytic function that does not satisfy a polynomial equation


However these definitions are arguably rather cryptic to those who are not familiar with the literature of higher mathematics.

So in layman's terms, what exactly does it mean to be transcendental? How would a transcendental number be different from an ordinary number, say 5. And respectively, how would a transcendental function be different from an ordinary function, say $f(x) = x^2$
We will play a game. Suppose you have some number $x$.  You start with $x$ and then you can add, subtract, multiply, or divide by any integer, except zero.  You can also multiply by $x$.  You can do these things as many times as you want. If the total becomes zero, you win.
For example, suppose $x$ is $\frac23$. Multiply by $3$, then subtract $2$. The result is zero. You win!
Suppose $x$ is $\sqrt[3] 7$. Multiply by $x$, then by $x$ again, then subtract $7$. You win!
Suppose $x$ is $\sqrt2 +\sqrt3$. Here it's not easy to see how to win. But it turns out that if you multiply by $x$, subtract 10, multiply by $x$ twice, and add $1$, then you win. (This is not supposed to be obvious; you can try it with your calculator.)
But if you start with $x=\pi$, you cannot win. There is no way to get from $\pi$ to $0$  if you add, subtract, multiply, or divide by integers, or multiply by $\pi$, no matter how many steps you take. (This is also not supposed to be obvious. It is a very tricky thing!)

Numbers like $\sqrt 2+ \sqrt 3$ from which you can win are called algebraic. Numbers like $\pi$ with which you can't win are called transcendental.

Why is this interesting?  Each algebraic number is related arithmetically to the integers, and the winning moves in the game show you how so.  The path to zero might be long and complicated, but each step is simple and there is a path. But transcendental numbers are fundamentally different: they are not arithmetically related to the integers via simple steps.
Well, in simple english which isn't rigorous, "transcendental" means it can not be expressed in terms of whole numbers and roots.  $\sqrt[3]{2 + 1/\sqrt{3}}, \sqrt[7]{43} + 5\sqrt[3]{49},$ etc (or simply $\sqrt[4]{3}$ or even $39/47$ or $5$) can be and so are not transcendental.  $\pi$ can not be expressed in any such way (take my word for it) so it is transcendental.

But you can see that is a terrible, poorly defined, ambiguous definition that is useless in any formal mathematical sense.

More formally a real number $m$ is algebraic if there is some polynomial $P(x) = a_nx^n + a_{n-1}x^{n-1} + .... + a_2x^2 + a_1x + a_0$ where each of the $a_i$s are integers and each of the $x^i$ are raised to a non-negative integer power-- if there is such a polynomial where if you plug in your number $m$ and get $P(m) = 0$-- if that is possible, the number is algebraic.  If it is not possible for there to be any such polynomial, the number is *transcendental.

So take my number $m =\sqrt[3]{2 + 1/\sqrt{3}}$ it is the solution to ... $P(x) = 3x^6-12x^3 + 11 = 0$[*].  There is no finite polynomial with integer coefficients where $P(\pi) = 0$.  (Take my word for that).

===

$3(\sqrt[3]{2 + 1/\sqrt{3}})^6 - 12 (\sqrt[3]{2 + 1/\sqrt{3}})^3 + 11 =$

$3({2 + 1/\sqrt{3}})^2 - 12({2 + 1/\sqrt{3}}) + 11 =$

$3(4 + 4/\sqrt{3} + 1/3) - 12(2 + 1/\sqrt{3}) + 11 =$

$12 + 12/\sqrt{3} + 1 - 24 - 12/\sqrt{3} + 11 = 0$

My other number  $\sqrt[7]{43} + 5\sqrt[3]{49}$ is also a solution to  polynomials but it'd be a real pain for me to figure out which ones.  (It'd be a 21 degree polynomial.  If I set $P(x) = a_{21}x^{21} + ... + a_1x + a_0$, plug in $ \sqrt[7]{43} + 5\sqrt[3]{49}$, set equal to 0, I will get 21 equations to solve for 22 of the $a_i$s in terms of the 22nd $a_i$ which I would choose just to make them all integers.  Very tedious but doable.  Always doable for any finite expression in terms of roots and whole numbers.)

$\sqrt[4]{3}$ is the solution to $x^4 - 3 = 0$.

$39/47$ is the solution to $47x - 39 = 0$ (which is a single degree polynomial) and $5$ is the solution to $x - 5 = 0$.