real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is 
Number of real solution of equation  $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is 

Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$ 
$x=-7,x=-1$
Solution given is all real solution
Help me please

Upon completing the square the equation $$(x^2+6x+7)^2+6(x^2+6x+7)+7=x$$
$$(x^2+6x+7)^2+6(x^2+6x+7)+9=x+2$$
$$[(x^2+6x+7)+3]^2=x+2$$
Which simplifies to $$[(x+3)^2+1]^2=x+2$$ which has no real solutions.
Upon comparison of derivatives for $x>-2$,  we see that  function on the LHS is always above the one on the RHS so they do not meet at any real values.
