Solving $\frac{x^{11}+x}{x^7+x^5}=\frac{205}{16}$ 
Blockquote
Find $x$ satisfy the equation$$\frac{x^{11}+x}{x^7+x^5}=\frac{205}{16}$$

Here is my work
$$\frac{x^{10}+1}{x^6+x^4}=\frac{205}{16},\quad\quad t:=x^2$$
$$\frac{t^5+1}{t^3+t^2}=\frac{205}{16}$$
$$16t^5-205t^3-205t^2+16=0$$
$t=-1$ is a root of the polynomial but not sure how to find the other roots.
Another approach I tried is dividing numerator and denominator of the original fraction by $x^6$,
$$\frac{x^5+\frac1{x^5}}{x+\frac1x}=\frac{205}{16}$$
$$x^4-x^2+1-\frac1{x^2}+\frac1{x^4}=\frac{205}{16}$$

As an alternative from here we have
$$\frac{t^5+1}{t^3+t^2}=\frac{205}{16} \iff \frac{(t+1)(t^4-t^3+t^2-t+1)}{t^2(t+1)}=\frac{205}{16}$$
and since $t=x^2\ge 0$
$$\iff \frac{t^4-t^3+t^2-t+1}{t^2}=\frac{205}{16} \iff t^2-t+1-\frac1 t+\frac1{t^2}=\frac{205}{16}$$
$$\iff t^2+2+\frac1{t^2}-t-\frac1 t-\frac{221}{16}=0 \iff \left(t+\frac1t\right)^2-\left(t+\frac1t\right)-\frac{221}{16}=0$$
and by $u= t+\frac1t$ we obtain

$$u^2-u-\frac{221}{16}=0$$

