Degrees And Roots Might Be Independent

The equation can be written as $f(x)=(x+5)^2-5$

now $f(f(x))=(x+5)^{ 4 }-5\\ f(f(f(x)))=(x+5)^{ 8 }-5\\ f(f(f(f(x))))=(x+5)^{ 16 }-5=p(x)$

It can easily be seen that $p(x)=0$ has $2$ real solns.

Aliter : One can even plot the graph.Its amazing!Really.

Note : We can show that $(x+5)^{ 16 }-5 = 0$ has only 2 real roots by considering plotting a graph of $y = x^{16}$ and a straight line $y=5$ .

$f\left( x \right) =a\quad ,f\left( f\left( x \right) \right) =f\left( a \right) =b\quad ,\quad f\left( f\left( f\left( x \right) \right) \right) =f\left( b \right) =c\quad \\ so\quad f\left( f\left( f\left( f\left( x \right) \right) \right) \right) =f\left( f\left( b \right) \right) =f\left( c \right) ={ c }^{ 2 }+10c+20=0\\ then\quad { c }^{ 2 }+10c+20=0\quad { c }_{ 1 }=-(5+\sqrt { 5 } )\quad ,{ c }_{ 2 }=\sqrt { 5 } -5\quad .\\ Then\quad we\quad solve\quad just\quad for\quad b\quad and\quad we\quad get\quad two\quad solutions\quad because\quad \\ the\quad equationts\quad have\quad the\quad same\quad result\quad \\ one\quad has\quad two\quad roots\quad the\quad other\quad has\quad no\quad real\quad roots\quad so\quad in\quad the\quad end\\ \quad we\quad have\quad 2\quad real\quad roots\quad { x }_{ 1 }{ =-(\sqrt [ 16 ]{ 5 } +5),\quad x }_{ 2 }=\sqrt [ 16 ]{ 5 } -5\quad \\$

Let c be a solution to the original expression. Therefore f(c)=f(f(f(f(x)))). Applying this method recursively you get 2 solutions.