Given one root is x1 = -2 + i.

Step 1: Identify the conjugate root. Since the polynomial $P(x) = x^4 + 4x^3 + 6x^2 + 4x + 5$ has real coefficients, if a complex number is a root, its complex conjugate must also be a root. Given one root is $x_1 = -2 + i$. Therefore, its conjugate $x_2 = -2 - i$ is also a root.

Step 2: Form a quadratic factor from these two roots. A quadratic factor corresponding to these roots is $(x - x_1)(x - x_2)$. $(x - (-2 + i))(x - (-2 - i))$ $= ((x+2) - i)((x+2) + i)$ This is in the form $(a-b)(a+b) = a^2 - b^2$, where $a = (x+2)$ and $b = i$. $= (x+2)^2 - i^2$ $= (x^2 + 4x + 4) - (-1)$ $= x^2 + 4x + 4 + 1$ $= x^2 + 4x + 5$

Step 3: Divide the original polynomial by this quadratic factor. We perform polynomial long division of $P(x) = x^4 + 4x^3 + 6x^2 + 4x + 5$ by $(x^2 + 4x + 5)$. $\begin{array}{c|cc cc cc} \multicolumn{2}{r}{x^2} & +0x & +1 \\ \cline{2-7} x^2+4x+5 & x^4 & +4x^3 & +6x^2 & +4x & +5 \\ \multicolumn{2}{r}{-(x^4} & +4x^3 & +5x^2) \\ \cline{2-4} \multicolumn{2}{r}{0} & 0 & x^2 & +4x & +5 \\ \multicolumn{2}{r}{} & \multicolumn{2}{r}{-(x^2} & +4x & +5) \\ \cline{4-6} \multicolumn{2}{r}{} & \multicolumn{2}{r}{0} & 0 & 0 \\ \end{array}$ The quotient is $x^2 + 1$. So, $P(x) = (x^2 + 4x + 5)(x^2 + 1)$.

Step 4: Find the roots of the remaining quadratic factor. Set the remaining factor equal to zero: $x^2 + 1 = 0$ $x^2 = -1$ $x = \pm\sqrt{-1}$ $x = \pm i$ So, the other two roots are $x_3 = i$ and $x_4 = -i$.

Step 5: List all the roots. The roots of the polynomial equation are $-2+i$, $-2-i$, $i$, and $-i$.

The roots of the polynomial equation are $\boxed{-2+i, -2-i, i, -i}$.

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