Answer
$-\infty \lt x \lt \infty$
The solution set is $~~(-\infty, \infty)$
Work Step by Step
$x^2+x+1 \gt 0$
We can use the quadratic formula to find the values of $x$ such that $x^2+x+1 = 0$:
$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$x = \frac{-1 \pm \sqrt{(1)^2-4(1)(1)}}{(2)(1)}$
$x = \frac{-1 \pm \sqrt{-3}}{2}$
Since there is a negative value inside the square root, there are no values of $x$ such that $x^2+x+1 = 0$
When $x = 1,~~$ then $~~x^2+x+1 = 3 \gt 0$
Since $x^2+x+1$ is continuous, and there are no values of $x$ such that $x^2+x+1 = 0$, then $x^2+x+1 \gt 0$ for all values of $x$
Solution:
$-\infty \lt x \lt \infty$
The solution set is $~~(-\infty, \infty)$
