That's Integral not Real!

Simplifying the given expression we get,

$x^2 - (a+10)x + 10a +1 = 0$

$\Rightarrow x = \dfrac{(a+10) \pm \sqrt{(a+10)^2 - 4(10a+1)}}{2}$

$\Rightarrow x = \dfrac{(a+10) \pm \sqrt{a^2 -20a + 96}}{2}$

$\Rightarrow x = \dfrac{(a+10)\pm \sqrt{(a-8)(a-12)}}{2}$

Let $a-12 =t$ , then,

$(a-12)(a-8) = t(t+4) = t^2 + 4t$ which is not a perfect for any $t \in \mathbb{Z} - \{0\}$ .

Hence the only chance that the roots could be integral is when $D = 0$

$(a-12)(a-8) = 0 \Rightarrow a = 12,8$

$x = \dfrac{(a+10)}{2}$ which is integral for $a=12 , 8$

Sum of values of $a$ is $\boxed{20}$ .