Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$. 
Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$.

My Attempt:
On rearranging, I get, $(x^2+x+1)(3x+7)+2=0$
Or, $3x^3+10x^2+10x+9=0$
Derivative of the cubic is $9x^2+20x+10$
It is zero at minus zero point something and minus one point something.
So, even at local minima, the cubic is positive. It means it would cross x-axis only once.
At $x=-2$, cubic is positive and at $-3$, it is negative.
It means the only root is minus two point something.
Is there any other way to solve this question? Something that doesn't involve calculator? Or maybe something that doesn't involve calculus?

$x^2+x+1 + \dfrac{2}{x^2+x+1} \ge 2\sqrt 2$
Largest value of $x^2 - 2x-6$ in the given interval is $2$ at $x=4$.
