How do you solve inequality ${{x}^{2}}+x+1>0$?

Hint: We first try to solve the equation ${{x}^{2}}+x+1=0$ to get the real solutions. Then we try to find the values of the curve for any values of x. We also find the minimum value of the function $f\left( x \right)={{x}^{2}}+x+1$. Then we find the solution of the inequality ${{x}^{2}}+x+1>0$.

Complete step-by-step solution:
We have to convert the given equation in such form so that we get the value of x directly.
We first try to find the real roots of the equation ${{x}^{2}}+x+1=0$ which will give us the intersection of the curve with the X-axis.
Using quadratic solving formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for $a{{x}^{2}}+bx+c=0$, we get $x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 1\times 1}}{2\times 1}=\dfrac{-1\pm i\sqrt{3}}{2}$ for ${{x}^{2}}+x+1=0$.
We don’t get any real solution. That means the curve never touches the X-axis. The curve is fully above or below the X-axis.
Now, \[{{x}^{2}}+x+1={{\left( x \right)}^{2}}+2\times x\times \dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}+1={{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}\].
The equation is the sum of a square value and a positive fraction \[\dfrac{3}{4}\].
Now we know that $\forall x\in \mathbb{R}$, value of \[{{\left( x+\dfrac{1}{2} \right)}^{2}}\ge 0\]. This means \[{{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}\ge \dfrac{3}{4}\].
We replace the function and get \[{{x}^{2}}+x+1\ge \dfrac{3}{4}\].
The value of the function is always greater than or equal to $\dfrac{3}{4}$.
This gives the solution of ${{x}^{2}}+x+1>0$ as $x\in \mathbb{R}$. Any value in real domain will solve the inequality.

Note: We can see that there is no maximum value for the function $f\left( x \right)={{x}^{2}}+x+1$. The range is $\dfrac{3}{4}\le {{x}^{2}}+x+1\le \infty $. The function is the factor of ${{x}^{3}}-1$. It gives the root of $\omega $ and ${{\omega }^{2}}$.