Is there a viable strategy to solve the following equation in an analytic way, without using numeric methods?
$(1+\frac{1}{8}x^2)^2=\frac{p^2}{2}(\sqrt{1+x^2}+1)$
Edit: When I substitute $t=\sqrt{1+x^2}$, I get:
$\frac{1}{64}(49+14t^2+t^4)=\frac{p^2}{2}(t+1)$
As you wrote, using $t=\sqrt{1+x^2}$, you end with $$t^4+14 t^2-32 p^2 t-(32 p^2-49)=0$$ which can be solved using Ferrari's method.
If you use the steps given here, you should find that $$\Delta=-1048576\, p^4 \left(p^2-1\right) \left(27 p^2+98\right)$$ which is negative if $p^2 \lt 1$.
We also get $P=112$, $Q=-256 p^2$, $\Delta_0=784-384 p^2$, $D=-2048 p^2$ and you can analyze the number and nature of roots depending on $p$. Now, play with the radicals (have fun) and come back to $x$ (more fun).
Personally, I would prefer to consider the function
$$f(x)=\frac{\left(1+\frac{x^2}{8}\right)^2}{2 \left(\sqrt{x^2+1}+1\right)}$$ which is symmetric and shows a minimum value equal to $\frac 14$ at $x=0$. This means that if $p^2 \lt \frac 14$ no root at all; if $p^2 = \frac 14$, a quadruple root equal to $0$ and if $p \gt \frac 14$, two real roots (good to know that they are symmetric) and two complex conjugate roots.
If you do not want to play with nasty radicals, just use a numerical method considering that you look for the zero of function $$g(y)=\frac{\left(1+\frac{y}{8}\right)^2}{2 \left(\sqrt{y+1}+1\right)}-p^2$$
If you consider Newton method, series expansion built at $x=0$ or $x\to \infty$ followed by series reversion will give quite good estimates.
