Let $2x^5+5x^4+10x^3+10x^2+10x+10=f\left(x\right)$

Now $f\left(-2\right)=-34$ and $f\left(-1\right)=3$

Hence $f\left(x\right)$ has a root in $\left(-2,-1\right)$

Further $f^{\text{'}}\left(x\right)=10x^4+20x^3+30x^2+20x+10$

$=10x^2\left(x^2+2x+3+\frac{2}{x}+\frac{1}{x^2}\right)$

$=10x^2\left[\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right)+3\right]$

$=10x^2\left[{\left(x+\frac{1}{x}\right)}^2-2+2\left(x+\frac{1}{x}\right)+3\right]$

$=10x^2\left[{\left(x+\frac{1}{x}+1\right)}^2\right]>0$  for all $x$ belongs to $R$.

$f\left(x\right)$ is strictly increasing function. Since it is an odd degree polynomial it will have exactly one real root.

Hence, $f\left(x\right)$ has only one real root, so $\left|a\right|=2$.