Four roots

Divide the given equation by $x^2$ , we get

$x^2 - 10x + 26 - 10 \frac{1}{x} + \frac{1}{ x^2 } = 0$

$[ x^2 + \frac{1}{ x^2 } ] - 10 [ x + \frac{1}{x} ] + 26 = 0$

Now,

Let, $x + \frac{1}{x} = y \Rightarrow x^2 + \frac{1}{ x^2 } = y^2 - 2$

$\Rightarrow (y^2 - 2) - 10y + 26 = 0$

$= y^2 - 10y + 24 = 0 \Rightarrow (y - 6) (y - 4) = 0$

$\Rightarrow y =$ 4 or 6

Now,

If $y = x + \frac{1}{x} = 6$

$\Rightarrow x^2 - 6x + 1 = 0$

$\Rightarrow x = \boxed{ 3 \pm 2 \sqrt{2} }$

If $y = x + \frac{1}{x} = 4$

$\Rightarrow x^2 - 4x + 1 = 0$

$\Rightarrow x = \boxed{2 \pm \sqrt{3} }$

So,

$a + b + c + d + e = \large \boxed{12}$

First of all lets consider the following expression:- $(x^{2} + \alpha x +1)^{2} = x^{4} + 2\alpha x^{3} + (\alpha^{2}+2)x^{2}+ 2\alpha x + 1$ If we put $\alpha=-5$ in the above equation we will get a similar expression as in the given equation. $(x^{2} - 5x+1)^{2} = x^{4} -10x^{3}+27x^{2} - 10x+1$ Hence the given equation can be written as:- $(x^{2} - 5x+1)^{2} - x^{2} =0$ $\implies (x^{2} -6x+1)(x^{2} -4x+1) =0$ Hence the expression has been factorized into two quadratic expression which can be solved using quadratic formula and hence we can reach our answer.