Solving $\log_2(x^2+ x +6 ) > 0$

Question

I have $$\log_2(x^2+ x +6 ) > 0$$

From the property of the $\log$ it should be the same as:

$$x^2+ x +6 > 2^0$$

therefore:

$$x^2+ x +5 > 0$$

But using the quadratic formula it gives me :

$$ x = \frac{-1\pm\sqrt{-19}}{2} =\frac{-1\pm i\sqrt{19}}{2} $$

From there I know that when the $x^2+ x +5 = 0$ but I do not know how can I find the solution in real numbers. It is the $$\frac{-1-i\sqrt{19}}{2}<x<\frac{-1+i\sqrt{-19}}{2}$$ or $$\frac{-1-\sqrt{-19}}{2}>x$$ and $$ x>\frac{-1+i\sqrt{-19}}{2}$$ ?

The result in real numbers after what WolframAlpha says should be $\frac{-1}{2}$, and intuitively I understand that. But I would not be able able to come up with it and the reason behind it.

How can I reason to find the real numbers solutions?

Answer 1

$$x^2+x+5=\left(x+\frac12\right)^2+\frac{19}{4}\ge\frac{19}4\gt0\quad\forall x\in\mathbb{R}$$

Answer 2

The minimum of $x^2+x+6$ is taking in $x=-0.5$ and is greater than $5$. Hence you do have your expression is always well defined (for all real).