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?

Expanding on Anne Bauval's comment.
Rearranged equation is $(x^2+x+1)(3x+7)+2=0$
Discriminant of $x^2+x+1$ is negative. Thus, this quadratic is always positive.
$-2\lt x\lt4\implies-6\lt3x\lt12\implies1\lt3x+7\lt19$
It means the rearranged equation, on the given interval is (positive)(positive)+positive. Thus, never zero.
So, the answer to the question is zero zeros in the interval.
