# How do you solve x^2+x+1>0 ?

Oct 2, 2015

#### Answer:

$x \in \mathbb{R}$

#### Explanation:

Consider the related equation ${x}^{2} + x + 1 = 0$

Evaluating the discriminant ($\Delta = {b}^{2} - 4 a c$)
$\textcolor{w h i t e}{\text{XXX}} \Delta = - 3$
and we know that if $\Delta < 0$ there are no Real solutions.

That is ${x}^{2} + x + 1$ does not touch or cross the X-axis.

We know that for some values (e.g. $x = 0$)
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + x + 1 > 0$
and since it doesn't touch or cross the X-axis
for all Real values of $x$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + x + 1 > 0$