Can you guess what I guessed!

The coefficients of the polynomial above hint at the equation being expressible in the form $(x^2-k)^2 = a(x-r)^2$ for constants $a,k,r$ . We just take the square root of both sides and Voila! We are done.

But, how do we do it? We can use trial and error to get $(x^2-5)^2 = 3(2x+1)^2$ , but how can we do an algebraic approach?

We can express the equation above in the form $x^4 -ax^2+(22+b) = (22-a)x^2+12x+b$ , where $a$ and $b$ are constants to be determined.

Now, in order for the LHS to be a perfect square trinomial in $x^2$ , we need $\frac{a^2}{4} = 22+b$ , or $4b = a^2 -88$ . Similarly, for the RHS to be a multiple of a perfect square trinomial, its discriminant must be equal to $0$ , or $144-(4b)(22-a) = 0$ , but by the equation above, $4b = a^2 -88$ , $144-(a^2 -88)(22-a) = 0$ .Solving for $a$ gives us $a=10$ , which consequently gives us $b = 3$ .

Then, we can get $x^4 -10x^2+25 = 3(4x^2+4x+1)$ , or similarly, $(x^2-5)^2 = 3(2x+1)^2$ , taking square root of both sides gives us

$x^2-5 = \pm 2x\sqrt{3} \pm \sqrt{3}$ , or equivalently, $x^2 \pm 2x\sqrt{3} -5 \pm \sqrt{3}=0$ . We then use the quadratic formula to obtain

$x = \pm \sqrt{3} \pm \sqrt{8 \pm \sqrt{3}}$ .

Therefore, $a = 3$ , $b = 8$ and $c = 3$ , giving us $a+b+c = \boxed{14}$