# X^4-10x^3+x^2-10x+1=0

`x^4-10x^3+x^2-10x+1=0`

Divide whole equation by `x^2` (we can do this because `x=0` is obviously not a solution).

`x^2-10x+1-10/x+1/x^2=0`

Now we make substitution `y=x+1/x` but first we regroup the terms of the equation.

`(x^2+2+1/x^2)+(-10x-10/x)-1=0`

`(x+1/x)^2-10(x+1/x)-1=0`

`y^2-10y-1=0`

Now we can use quadratic formula.

`y_(1,2)=(10pmsqrt(100+4))/2`

`y_1=(10-2sqrt26)/2=5-sqrt26`

`y_2=5+sqrt26`

For `y_2` we get

`x+1/x=5+sqrt26`

`(x^2+1)/x=5+sqrt26`  ` `

Multiply by `x.`

`x^2-(5+sqrt26)x+1=0`

`x_(1,2)=(5+sqrt26pmsqrt((5+sqrt26)^2-4))/2=`

`(5+sqrt26pmsqrt(47+10sqrt26))/2`

`x_1=(5+sqrt26-sqrt(47+10sqrt26))/2`

`x_2=(5+sqrt26+sqrt(47+10sqrt26))/2`

`x_1` and `x_2` are first two solutions.

Now we put `y_1` into our substitution.

`x+1/x=5-sqrt26`

`(x^2+1)/2=5-sqrt26`

`x^2-(5-sqrt26)x+1`

`x_(3,4)=(5-sqrt26pmsqrt(47-10sqrt26))/2`

`x_3=(5-sqrt26-i sqrt(10sqrt26-47))/2` <-- Third solution

`x_4=(5-sqrt26+isqrt(10sqrt26-47))/2`  <-- Fourth solution

Solutions of the equation are `x_1,x_2,x_3,x_4.`