Let $P(x)=2x^5+5x^4+10x^3+10x^2+10x+10.$ We need an integer $a$ such that all real roots of $P(x)$ lie in $(a,a+1)$.

Rewrite the polynomial in a useful form. Using $(x+1)^5=x^5+5x^4+10x^3+10x^2+5x+1,$ we get $2(x+1)^5=2x^5+10x^4+20x^3+20x^2+10x+2.$ This is not exactly $P(x)$, so instead observe: P(x)=(x+1)^5+igl(x^5+5x^3+5x^2+5x+9igr), which is not especially helpful. So let us analyze roots directly.

First, determine how many real roots $P(x)$ can have. Differentiate: $P'(x)=10x^4+20x^3+30x^2+20x+10=10(x^4+2x^3+3x^2+2x+1).$ Now factor: $x^4+2x^3+3x^2+2x+1=(x^2+1)(x^2+2x+1)=(x^2+1)(x+1)^2.$ Hence, $P'(x)=10(x^2+1)(x+1)^2\ge 0,$ and in fact $P'(x)=0$ only at $x=-1$. Therefore $P(x)$ is an increasing function on $\mathbb R$ (non-decreasing everywhere, strictly increasing except for a stationary point at $x=-1$). So $P(x)$ can have at most one real root.

Now locate that real root by checking values. Compute: $P(-2)=2(-32)+5(16)+10(-8)+10(4)+10(-2)+10$ $=-64+80-80+40-20+10=-34<0.$ Next, $P(-1)=2(-1)+5(1)+10(-1)+10(1)+10(-1)+10$ $=-2+5-10+10-10+10=3>0.$ Since $P(x)$ is continuous, there is a real root in $(-2,-1)$.

Because $P(x)$ has at most one real root and we have found one in $(-2,-1)$, this is the only real root, and hence all real roots lie in the interval $(-2,-1).$ Thus, $a=-2.$ Therefore, $|a|=2.$

Compare with the stored correct answer. Stored correct answer = $2$, which matches our result.