The idea is basically:
Any monic polynomial can be factored as $f(x) = \prod (x - a_i)$, where $a_{1,\dots,n}$ are the roots of the polynomial.
Now if we expand such a product:
$(x - a_1)(x - a_2) = x^2 - (a_1 + a_2)x + a_1a_2$
$(x - a_1)(x - a_2)(x - a_3) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_1a_3 + a_2a_3)x - a_1a_2a_3$
And so on. The pattern should be clear.
This means that finding the roots of a polynomial is in fact equivalent to solving systems like the following:
For a quadratic polynomial $x^2 - px + q$, find $a_1,a_2$, such that
$p = a_1 + a_2$
$q = a_1a_2$
For a cubic polynomial $x^3 - px^2 + qx - r$, find $a_1,a_2,a_3$, such that
$p = a_1 + a_2 + a_3$
$q = a_1a_2 + a_1a_3 + a_2a_3$
$r = a_1 a_2 a_3$
And similarly for higher degree polynomials.
Not surprisingly, the amount of "unfolding" that needs to be done to solve the quadratic system is much less than the amount of "unfolding" needed for the cubic system.
The reason why polynomials of degree 5 or higher are not solvable by radicals, can be thought of as: The structure (symmetries) of the system for such a polynomial just doesn't match any of the structures that can be obtained by combining the structures of the elementary operations (adding subtracting, multiplication, division, and taking roots).
The equation you want to develop being $$\frac{\sqrt{x^2-9}+3\sqrt{x^2-1}}{x^2}-\frac12\sqrt{3}=0$$ start setting $x^2=t$ as suggested by Shubham; so it becomes $$\frac{\sqrt{t-9}+3\sqrt{t-1}}{t}-\frac12\sqrt{3}=0$$ Multiply everything by $t$ and move the second part to the rhs. The expression becomes $$\sqrt{t-9}+3\sqrt{t-1}=\frac {\sqrt{3}}{2}t$$ Let us define $a=\sqrt{t-9}$,$b=3\sqrt{t-1}$,$c=\frac {\sqrt{3}}{2}t$. So we have now $$a+b=c$$ Square both sides and develop, moving $a^2$ and $b^2$ to the rhs. So $$2ab=c^2-a^2-b^2$$ and square again for reaching finally $$4a^2b^2=(c^2-a^2-b^2)^2$$ and now replace. Before simplification, the result is $$-\frac{9 t^4}{16}+15 t^3-91 t^2=0$$ $t^2$ can be factored and you end with $$(-\frac{9 t^2}{16}+15 t-91) t^2=0$$ The roots of the quadratic equation are $t=\frac{28}{3}$ and $t=\frac{52}{3}$. We must discard $t=0$ and $t=\frac{28}{3}$ because of the domain definition. So, the only root left is $t=\frac{52}{3}$ and, since $t=x^2$, the solutions are $$x_{\pm}=\pm \sqrt \frac{52}{3}$$ Just be patient.
Best Answer
Great question I believe! You probably know this but this is said in MacTutor's Quadratic, cubic and quartic equation.
I just wanted to put it here to emphasize how "non-trivial" the problem of solving quartic equations can be. (I believe that you are aware of methods of finding rational roots, Uspensky's Theory of Equations is a great reference for that; but the general problem is definitely non trivial).
Here are many of this methods explained. I also found some of them here.