<s>[INST] Ladder against a wall. Having a bit of a problem with a question. There is a 4m ladder leaving against a wall. There is a box in between The ladder and wall. The box is a cubic metre.
I have found a quartic to find the length up the wall the ladder will reach. But its proving difficult for me to solve. 
$$x^4 +2x^3 -14x^2 +2x +1=0$$
 [/INST] You can find more about this type of equations under the name reciprocal equation or reciprocal polynomial.
See, for example, also this post Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ (And several of the posts shown there among linked questions.)

In this particular case, you have:
$$
\begin{align}
x^4+2x^3-14x^2+2x+1&=0\\
x^2+2x-14+\frac2x+\frac1{x^2}&=0
\end{align}
$$
If we use the substitution $u=x+\frac1x$, we get $u^2=x^2+2+\frac1{x^2}$ $\Rightarrow$ $x^2+\frac1{x^2}=u^2-2$. So your equation gets to the form
$$
\begin{align}
\left(x^2+\frac1{x^2}\right)+2\left(x+\frac1x\right)-14&=0\\
u^2+2u-16=0\\
(u+1)^2-17=0
\end{align}
$$
which yields
$$u_{1,2}=-1\pm\sqrt{17}.$$
Now we have to solve for each $u$ the equation
$$
\begin{align}
x+\frac1x=u\\
x^2-ux+1=0
\end{align}
$$
which yields
$$x=\frac{u\pm\sqrt{u^2-4}}2.$$
Since we have $u^2=16-2u$, this can be rewritten as
$$x=\frac{u\pm\sqrt{12-2u}}2.$$
So the solutions are
$$
\begin{align}
x=\frac{-1\pm\sqrt{17}+\sqrt{14\mp2\sqrt{17}}}2
\end{align}
$$
Here is what WolframAlpha returns for your equation: (Link)
Note that you can switch there between approximate and exact forms of the result.
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