Find all four roots of quartic equation $x^4-x+1=0$ How to solve
$$x^4-x+1=0$$
My attempt:
$$x^4-x+1=0$$
$$\implies x^4-x^3-x+1+x^3=0$$
$$\implies x^3(x-1)-(x-1)+x^3=0$$
$$\implies (x^3-1)(x-1)+x^3=0$$
But, I couldn't find a way to combine $x^3$ into that factorization.
I then looked at Wikipedia to see how to solve a quartic. I'm not sure which method is the best one. The coefficients are pretty simple (in the form $ax^4+bx^3+cx^2+dx+e$, $a=e=1$, $b=c=0$, $d=-1$). Should I just use the general formula for quartic equations, or something else?
Also, I couldn't find a post on here talking about how to solve quartic equations. If someone finds a link to such a post then I might as well just delete this question.
The only post I found that might be useful is this question but sadly there are no answers there.
EDIT: I would prefer all four solutions, real or complex.
 A: There are no real solutions because $x^4-x+1$ attains a positive minimum at $x=1/\sqrt[3]{4}$.
A: Note that $x^4-x+1=0$ is a deeply-depressed quartic equation, which makes it manageable. In fact, it can be factorized as
$$x^4-x+1= \left( x^2- ax+ \frac{a^3-1}{2a} \right)  \left( x^2+ ax+ \frac{a^3+1}{2a} \right) =0\tag1
$$
where $a$ satisfies the cubic equation $(a^2)^3-4a^2-1=0$ and can be obtained analytically
$$a = \sqrt{\frac4{\sqrt3} \cos\left( \frac13\cos^{-1}\frac{3\sqrt3}{16}\right)}$$
Then, solve the two quadratic equations in (1) to obtain the four complex roots
$$x = \frac a2 \pm \frac i2\sqrt{a^2-\frac2a},\>\>\>
-\frac a2 \pm \frac i2\sqrt{a^2+\frac2a}
$$
A: A new method for solving quartics known as the ferrari method which has quite posts on this site
so we add a factor of $(ex+f)^2$ on both sides so the equation becomes 
$$(x^2+ax+b)^2=(ex+f)^2$$
and we have to determine $a,b,e,f$

so expand $(x^2+ax+b)^2$ and you will get
  $$x^4+a^2x^2+b^2+2bx^2+2ax^3+2abx=x^4-x+1+e^2x^2+f^2+2efx$$
  on comparing coefficients we get $$\begin{align} a =0 \rightarrow (1)
& \\2ef = 1 \ \ \  \rightarrow (2) \\1+f^2=b^2\rightarrow (3) \\e^2 = 2b\rightarrow (4) \end{align}$$
  now square the $2^{nd}$ equation to get $$f^2 = \frac{1}{8b}$$
  put this result in $(3)$ and form a cubic polynomial in $b$
  which is 
  $$8b^3-1-8b=0$$
  after this I think you can proceed

A: It's sufficent to show that it has no roots in $\mathbb{R}$:
Let $f(x)=x^4-x+1$, then $f'(x)=4x^3-1$, $x_0=\sqrt[3]{\frac{1}{4}}$,
$f(x)$ decreases on $(-\infty,x_0)$ and increases on $(x_0,\infty)$ so it's sufficent to find $f(x_0)$.
$$f(x_0)=\frac{1}{8}\left(8-3\sqrt[3]{2}\right)>0\hbox{ as }
8^3>3^3\cdot 2$$
For complex roots one can try Ferrari method. Encyclopedia of Mathematics.
A: Before diving into any details, I consulted Wolfram Alpha and noted that the roots are non-real complex conjugate pairs. Results from Wolfram Alpha for $x^4-x=1=0$.
From the section on the nature of solutions, I cite:
The possible cases for the nature of the roots are as follows: [...] If $P > 0$ or $D > 0$ then there are two pairs of non-real complex conjugate roots. [...]
We calculate some of the related coefficients. We find that $$P=8ac-3b^2=0$$ and $$R=b^3+8da^2-4abc=-8<0$$ and $$D=64a^3e-16a^2c^2+16ab^2c-16a^bd-3b^4=64>0$$ and  $\Delta_0=12>0$. 
The case $P=0$ and $D>0$ does not seem to be listed. But actually, I should have started with the discriminant $\Delta$ (which has only two non-zero terms, subject to human error) and I calculate that $\Delta=229>0$. 
Whenever $\Delta>0$, all four roots are real or none of them are. I do not see a reason why $P=0$ is not listed. 
A: The Newton-Raphson method uses an iterative process to approach one root of a any function:
$$x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)}$$
This could be a method to see that the equation $$x^4-x+1=0$$ not have any real solutions.
Indeed if you separated the fourth degree equation into two functions, the first $f(x)=x^4$ and the second $g(x)=x-1$, starting from the equation $x^4-x+1=0$, using Desmos to draw one function, for example, you can observe that there is no intersections beetween $f$ and $g$. I have chosen the graphic way.

A: Equation $\displaystyle p x + x^4 = t$
Solution:
$\displaystyle Q = ((-(27 p^4 + 128 t^3) + 3 (3 p^4 (27 p^4 + 256 t^3))^{1/2})/2)^{1/3}$
$\displaystyle A = (Q + 4 t (4 t/Q - 1))/(6 p)$
$\displaystyle B = (32 (3 p A + t))^{-1/6}$
$\displaystyle F = 256 B^{12} t (16 A^4 + 2 A p - t)$
$\displaystyle R_2 = cos((arccos(1 + 8 F) + 2 \pi j)/4)$
$\displaystyle j=0,1,2,3$
$\displaystyle R = 4 B^3 (1 - A^2)$
$\displaystyle y = (R_2 - R)/(4 B^4)$
$\displaystyle x= A \pm (1 + B y)^{1/2}$
A: The polynomial is irreducible but solvable.
$$x≈-0.72714 \pm 0.93410 i\qquad \land\qquad x≈0.72714 \pm 0.43001 i$$
