Solving the polynomial equation: $1+x^2-\frac{t}{x^{d-2}}=0$ I was attempting to solve the following equation for $x$:
$$1+x^2-\frac{t}{x^{d-2}}=0$$
where $t\in\mathbb{R}_{+}$ and $d\in \mathbb{Z_{+}}$.
Is it possible to obtain a general solution for such an equation? 
 A: As already pointed out, this equation is essentially equivalent to
$$
x^d + x^{d-2} - t = 0
$$
We may notice that for $d=2,3,4$ we have an equation of degree at most 4, that thus has an explicit solution. In particular, in the case $d=4$ the expression for the solutions is relatively easy since we may set $y=x^2$ and solve the quadratic equation $y^2+y-t=0$.
If $d=6$ or $d=8$ again we may obtain an explicit expression for the solutions setting $x^2=y$ and solving the cubic and quartic resulting equations.
For $d=5$ we have a quintic equation $x^5+x^3-t=0$. 
We know that in general we cannot express the solutions using the 4 basic arithmetic operations and some roots.
The same holds, for $d=7,9,10,11,12,...$.
However, if we do not impose such a severe restriction on the functions or operations we may use to express the roots, there is actually a way (albeit a very complicated one).
According to MathWorld, the roots of a quintic can be expressed in terms of Jacobi Theta functions.
The same MathWorld page also says that in general the roots of equations of the form
$$x^p+b\cdot x^q+c=0$$
can be expressed in terms of hypergeometric functions.
Clearly your equation is of this form, so in general its root could be expressed as well in terms of hypergeometric functions, but probably this is not going to be easy. You could check the refs in that MathWorld page if you are interested pursuing this avenue.
