Argument of Dirac delta function is quadratic with complex roots
I shall provide more details if I am correct. IMO if there is not root on the domain of integration, and here I suppose that $t\in \mathbb{R}$, then the argument never vanishes. Then, in an engineer fashion, one could say (but $t\mapsto \delta(t)$ is not a function): 

$$\delta(t^2+t+1) \equiv 0\,.$$

In more precise words, I would consider that for any suitable $f(t)$, $t\in \mathbb{R}$:

$$\int_{-\infty}^{\infty}f(t)\delta(t^2+t+1)dt = 0\,.$$

Related links:


Dirac $δ(g(x))$ with complex roots $x_i $
Regarding $x^2−a^2$ inside the argument of Dirac delta