By the Vieta's theorem, we have $$(x_1+1)(x_2+1)=x_1x_2+(x_1+x_2)+1=(k^2+2) + 2(k+1)+1 =k^2 + 2k+5$$ Therefore, $$k^2 + 2k+5 = 8\implies k^2 +2k-3=0 \implies (k+3)(k-1)=0\implies k = 1, -3$$ Meanwhile, in order for this equation to be solvable in real numbers, its determinant must be non-negative: $$(2(k+1))^2 - 4\cdot(k^2+2)\ge 0\implies 2k -1 \ge 0\implies k\ge\frac{1}{2}$$ Hence, the final answer is $\boxed{1}$.