$h^2 = x^2 + (x+1)^4$

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This is a question I had of finding an exact value of $h$ as I was interested in the process and techniques used. The question was one to do with Pythagoras and the math shown is the question in $a^2 + b^2 = c^2$ form.

The question was what is the size of the hypotenuse of a right-angled triangle if one side is a size $x$ and the other is a size $x^2 + 2x + 1$ (which was rewritten as $(x + 1)^2$). I got an answer from Wolfram Alpha of a value for $h$ that does not include $x$, only numbers.

Wolfram Input:

Wolfram Alpha Input

Wolfram Answer: Wolfram Alpha Real Solution

What would be the process of finding the exact value of h as to where there is no sign of x in the h = equation?

P.S. If the process is simply too lengthy or tedious, feel free to take this question down.

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\begin{align} c^2 &= a^2 + b^2 \\[1em] h^2 &= x^2 + \left[(x+1)^2\right]^2 \\ h^2 &= x^2 + (x+1)^4 \\ h^2 &= x^2 + x^4 + 4x^3 + 6x^2 + 4x +1\\ h^2 &= x^4 + 4x^3 + 7x^2 + 4x +1\\ h &= \sqrt{x^4 + 4x^3 + 7x^2 + 4x +1} \end{align}

You probably put into Wolfram Alpha some other task - you wanted the square root, and you obtained the root (= solution) of the system of equations with unknowns $h$ and $x$.

Addendum:

If you wanted from Wolfram Alpha the result of my computing, enter into it

solve(h^2 = x^2 + (x+1)^4, h)

(h at the end determines which you want to compute), and you will obtain

$$h = \pm \sqrt{x^4 + 4x^3 + 7x^2 + 4x +1}$$

If you want the result for a particular $x$, enter its value, e. g.

h^2 = x^2 + (x+1)^4, x=1

to obtain plot

enter image description here

and solutions $$\begin{array}{l}{h=-\sqrt{17}, \quad x=1} \\ {h=\sqrt{17}, \quad x=1}\end{array}$$