They are a and b

Solving for $b$ from the given equation, we get $b=\frac{-(a-1)^2\pm \sqrt{-(1+a)^2 (7a^2+2a+7)}}{2(2+a+a^2)}$ Now, $\displaystyle 7a^2+2a+7=7\bigg[\Big(a+\frac{1}{7}\Big)^2+\frac{48}{49}\bigg]>0 \:(\text{as } a \text{ is real}) \\ \implies -(1+a)^2 (7a^2+2a+7)\leq0$

So, $b$ is real iff $(1+a)^2 =0 \implies a=-1$ .

Using the solution of $b$ we get $b=-1\implies \boxed{a+b=-2}$

Try to find symmetry solution a=b, it is simple a=-1 and a+b=2a=-2.