<s>[INST] How to solve $2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0$? 
How to solve $$2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0?$$

I just have no idea and I'have some knowledge about the polynomial equations. Here, just nothing. 
 [/INST] We had one root $x=-\frac{1}{\sqrt{2}}$ as already said in answers and comments. After long division remains $$2 x^4+4 \sqrt{2} x^3+16 x^2+12 \sqrt{2} x+8=0$$
Make $x=y-\frac{1}{\sqrt{2}}$ to get
$$2 y^4+10 y^2+\frac{5}{2}=0$$ which is a quadratic in $y^2$.
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