Roots of the equation $(x^2+3x+4)^2+3(x^2+3x+4)+4=x$ The equation $(x^2+3x+4)^2+3(x^2+3x+4)+4=x$ has
(A) all its solution real but not all positive
(B) only two of its solution real
(C) two of its solution positive and two negative
(D) none of its solution real
My approach is as follow $(x^2+3x+4)^2+3(x^2+3x+4)+4-x=y$
$f(x)=y=x^4+6x^3+20x^2+32x+32$
$f(-x)=y=x^4-6x^3+20x^2-32x+32$
Using Descartes rule no positive roots but the possible ways we can have either 4,2,0 negative roots.
$f'(x)=4x^3+18x^2+40x+32$
$f''(x)=12x^2+36x+40$ which is imaginary
From here I am not able to approach. 
 A: Let $t=x^2+2x+4$, then we can write the equation as:
$$(t+x)^2+3(t+x)+4-x=0$$
or 
$$t^2+2tx+x^2+3t+2x+4=0$$
or
$$t^2+2tx+3t+t=0$$
or
$$t(t+2x+4)=0$$
so
$$(x^2+2x+4)(x^2+4x+8)=0$$
Can you end it now?
A: Hint (edited): Let $f(x) = x^2+3x+4$. Then $f(x) > x $ for all reals $x$, hence $f(f(x)) > f(x) > x$ for all reals $x$.
A: $$(x^2+3x+4)^2+3(x^2+3x+4)+4=x$$
$$(x^2+3x+4)^2+4(x^2+3x+4)+4=x+x^2+3x+4$$
$$(x^2+3x+4+2)^2=(x+2)^2$$
$$(x^2+3x+6)^2-(x+2)^2=0$$
then use the fact
$$a^2-b^2=(a-b)(a+b)$$
A: $\displaystyle f(x)=(x^2+3x+4)^2+3(x^2+3x+4)+4-x$
$\displaystyle f(x)=\bigg[\bigg(x+\frac{3}{2}\bigg)^2+\frac{7}{4}\bigg]^2+3\bigg[\bigg(x+\frac{3}{2}\bigg)^2+\frac{7}{4}\bigg]+4-x$
put $\displaystyle x+\frac{3}{2}=t\in \mathbb{R}$
$\displaystyle f(t)=\bigg(t^2+\frac{7}{4}\bigg)^2+3\bigg(t^2+\frac{7}{4}\bigg)+4-t+\frac{3}{2}$
$\displaystyle f(t)=t^4+\frac{49}{16}+\frac{7}{2}t^2+3t^2+\frac{21}{4}+4+\frac{3}{2}-t$
$\displaystyle f(t)=t^4+\frac{13}{2}\bigg[t^2-\frac{2}{13}t+\frac{1}{13^2}+\frac{43}{26}-\frac{1}{13^2}\bigg]$
$\displaystyle f(t)=t^4+\frac{13}{2}\bigg[\bigg(t-\frac{1}{13}\bigg)^2+\frac{43\cdot 13-2}{26\cdot 13}\bigg]>0$
