How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have? 
How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?

I don't know how to start...

Let's start by applying the quadratic formula.
We have: $(x^{2}+4x+8)^{2}+3x\hspace{1 mm}(x^{2}+4x+8)+2x^{2}=0$
$\Rightarrow x^{2}+4x+8=\dfrac{-3x\pm\sqrt{(3x)^{2}-4(1)(2x^{2})}}{2(1)}$
$\hspace{27.75 mm} =\dfrac{-3x\pm{x}}{2}$
$\hspace{27.75 mm} =-2x,-x$
Then,
$\Rightarrow x^{2}+4x+8=-2x$
$\Rightarrow x^{2}+6x+8=0$
$\Rightarrow x^{2}+2x+4x+8=0$
$\Rightarrow x\hspace{1 mm}(x+2)+4\hspace{1 mm}(x+2)=0$
$\Rightarrow (x+2)(x+4)=0$
$\Rightarrow x=-2,-4$
or
$\Rightarrow x^{2}+4x+8=-x$
$\Rightarrow x^{2}+5x+8=0$
$\Rightarrow x=\dfrac{-5\pm\sqrt{5^{2}-4(1)(8)}}{2(1)}$
$\hspace{9 mm} =\dfrac{-5\pm\sqrt{-7}}{2}$
$\hspace{9 mm} =\dfrac{-5\pm\sqrt{7}i}{2}$
$\hspace{9 mm} =\dfrac{-5-\sqrt{7}i}{2},\dfrac{-5+\sqrt{7}i}{2}$
However, $x\in\mathbb{Q}$.
Therefore, the solutions to the equation are $x=-2$ and $x=-4$.
