Solve the equation 

$2x^5+x^4-12-12x^2+x+2=0$

Given 

$2x^5+x^4-12-12x^2+x+2=0$

The equation is an odd degree reciprocal

equation of close one

$\therefore$$\text{'}-1\text{'}$ is a root of this equation 

$2x^4-x^3-11x^2-x+2=0$

It is an even degree reciprocal equation of class one. dividing with $\text{'}{x}^2\text{'}$

$$\begin{gathered} \frac{2x^4}{x^2}-\frac{x^3}{x^2}-\frac{11x^2}{x^2}-\frac{x}{x^2}+\frac{2}{x^2}=0 \\ 2x^2-x-11-\frac{1}{x}+\frac{2}{x^2}=0 \\ 2\left(x^2-\frac{1}{x}\right)-\left(x+\frac{1}{x}\right)-11=0\to \left(1\right) \\ Put x+\frac{1}{x}=y S.O.B.S \\ {x}^2+\frac{1}{x^2}+2x\frac{1}{x}=y^2 \\ {x}^2+\frac{1}{x^2}=y^2-2 \\ from \left(1\right) 2\left(y^2-2\right)-y-11=0 \\ 2y^2-y-15=0 \\ 2y^2-6y+5y-15=0 \\ \left(y-3\right)\left(2y+5\right)=0 \\ y-3=0 or \left(2y+5\right)=0 \\ y=3 \left(or\right) 2y+5=0 \\ case-i If y=3\Rightarrow x+\frac{1}{x}=3 \\ \Rightarrow x^2+1=3x \\ {x}^2-3x+1=0 \\ x=\frac{-\left(3\right)\pm \sqrt{9-4\left(1\right)\left(1\right)}}{2\left(1\right)}=\frac{3\pm \sqrt{5}}{2} \\ Case-ii: If y=\frac{-5}{2}\Rightarrow x=\frac{1}{x}=\frac{-5}{2} \\ x+\frac{1}{x}=\frac{-5}{2}\Rightarrow x^2+1=\frac{-5x}{2} \end{gathered}$$

$$\begin{gathered} 2x^2+5x+2=0 \\ \left(x+2\right)\left(2x+1\right)=0 \\ x+2=0 x=-2 \\ 2x+1=0 x=-1/2 \\ The roots are \frac{3\pm \sqrt{5}}{2}, -2, -1/2, -1 \end{gathered}$$