All the powers of 2 are even so this 6-degree expression can be reduced to a cubic if we put y=x^2:
2y^3+5y^2-y. Putting this equal to zero to find the y zeroes: 2y^3+5y^2-y=0. This factorises: y(2y^2+5y-1)=0. Take the quadratic: 2y^2+5y-1=0 and rewrite it as: y^2+5y/2=1/2 by dividing through by 2 and moving the constant across.
Now we can complete the square by dividing the y term by 2 and squaring its coefficient:
y^2+5y/2+25/16=1/2+25/16, and you can see we've added 25/16 to both sides of the equation to make it balance. We now have a perfect square on the left:
(y+5/4)^2=1/2+25/16=(8+25)/16=33/16, so we can take the square root of each side:
(y+5/4)=+sqrt(33)/4. Therefore, y=-5/4+sqrt(33)/4. This is the same result as we would have found using the formula for solving a quadratic equation. We need to calculate what these solutions are: 0.18614 and -2.68614. But we want x, not y, so we need to find the square root of these values of y. Assuming we don't want complex solutions (involving the imaginary square root of -1), we can only use the positive solution 0.18614 and take the square root of that, which is +0.43144 approx. so the real solution is x=0.43144 or -0.43144. (The complex solution is +1.63894i, where i=sqrt(-1).)
But we're not finished yet, because y=0 was also a solution, and that means x=0 is a solution, so we have three possible real zeroes for x: 0, 0.43144 and -0.43144 (and two complex zeroes).