Solving a 6th degree polynomial equation

The Rational Root Test shows that the only possible rational solutions are $\pm 1$. Substituting gives that $x = -1$ is one (but $x = 1$ is not), so polynomial long division gives $p(x) = -(x + 1) q(x)$ for some quintic $q$. Substituting $x = -1$ gives that $-1$ is not a root of $q$, so if $q$ factors over $\Bbb Q$, it does so into an irreducible quadratic and an irreducible cubic. One can find such a factorization without too much effort (this is made easier by the fact that the leading and constant coefficients are both $1$): We get $$p(x) = -(x + 1)\underbrace{(x^2 - x + 1)(x^3 - x^2 - 2 x + 1)}_{q(x)} .$$ The discriminant of the quadratic is $-3 < 0$, so the real root you've identified must be a factor of the cubic; since the cubic has no rational roots, one needs to use Cardano's Formula or the equivalent to extract it.

Using a complicated computer algebra system (Mathematica 10.4), we can get all the roots to this equation as radicals. Two roots are complex and the rest are all real (surprisingly).

Module[{roots},
  roots = Solve[-x^6 + x^5 + 2 x^4 - 2 x^3 + x^2 + 2 x - 1 == 0, x];
  Transpose[{
    N[x /. roots],
    FullSimplify[Element[x, Reals] /. roots],
    x /. roots
  }]
] // TableForm

• There are two complex roots at $\frac{1}{2} \pm \mathrm{i}\sqrt{3}$.
• There is a real root at $-1$.
• The other three are complicated and real: \begin{align} 1.80194\dots{} &= \frac{1}{3} \left(1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(-1+3 \mathrm{i} \sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(-1+3 \mathrm{i} \sqrt{3}\right)}\right), \\ -1.24698\dots{} &= \frac{1}{3}-\frac{7^{2/3} \left(1+\mathrm{i} \sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{-1+3 \mathrm{i} \sqrt{3}}}-\frac{1}{6} \left(1-\mathrm{i} \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 \mathrm{i} \sqrt{3}\right)}, \\ 0.445042\dots{} &= \frac{1}{3}-\frac{7^{2/3} \left(1-\mathrm{i} \sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{-1+3 \mathrm{i} \sqrt{3}}}-\frac{1}{6} \left(1+\mathrm{i} \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 \mathrm{i} \sqrt{3}\right)} \text{.} \end{align}

Conveniently, the first one is in the interval you require.

These three complicated roots are the roots of the same polynomial @Travis gets.

Looking at the Galois group structure, I believe we cannot dispense with complex numbers in these expressions.