One solution means a lot!

A very important property that one should know before proceeding with the problem

If $\log_cd > \log_cf$ and $d < f$ , then it can be easily inferred that the base $c$ is less than $1$ .

Substituting $x = \dfrac{4}{9}$ in the inequation, we get

$\log_a\dfrac{142}{81} > \log_a\dfrac{299}{81}$

$\Rightarrow a \in (0,1)$

So when we remove the logarithm the inequality is reversed.

$x^2 - x + 2 < -x^2 + 2x + 3 \Rightarrow 2x^2 - 3x - 1 < 0$

The solution set to the above inequality is $x \in \left( \dfrac{3-\sqrt{17}}{4} , \dfrac{3+\sqrt{17}}{4} \right) \ldots (1)$

But we simply cannot conclude this is possible without checking the domains of the logarithms in the R.H.S and L.H.S

For the R.H.S logarithm to exist,

$-x^2 + 2x + 3 > 0 \Rightarrow x^2 - 2x - 3 < 0 \Rightarrow x \in (-1,3) \ldots (2)$

For the L.H.S logarithm to exist,

$x^2 - x + 2 > 0 \Rightarrow \left(x-\dfrac{1}{2}\right)^2 + \dfrac{7}{4} > 0 \Rightarrow x \in \mathbb{R} \ldots (3)$

Considering the intersection of $(1),(2) \text{ \& } (3)$ , we get

$x \in \left( \dfrac{3-\sqrt{17}}{4} , \dfrac{3+\sqrt{17}}{4} \right) \Rightarrow x \in (-0.280 , 1.780)$

The integral values of $x$ in the given interval are $\boxed{0},\boxed{1}$

Sum of the values is $\large \color{#D61F06}{\boxed{1}}$