Question 1104188
There has to be a better way, but here it goes.
If f[g(x)] ={{{4x^2 + 10x + 5}}}
{{{g(x)}}}={{{f^-1}}}{ f[g(x)] } ={{{f^-1(4x^2 + 10x + 5)}}}
To find the inverse function of {{{f(x)=x^2-x-1}}} , {{{graph(300,300,-5,5,-2,8,x^2-x-1)}}} ,
we exchange {{{x}}} and {{{y}}} in
{{{y=x^2-x-1}}} to get {{{x=y^2-y-1}}} and then solve for {{{y}}} .
We get {{{y=(1 +- sqrt(4x+5))/2}}} , {{{graph(300,300,-5,5,-5,5,(1+sqrt(4x+5))/2,(1-sqrt(4x+5))/2)}}} .
That is really two functions,
and when working with real numbers,
it is only defined for {{{x>=-5/4}}} ,
but we continue, to see what we can get.
If {{{f^-1(x)=(1 + sqrt(4x+5))/2}}} 
{{{f^-1(4x^2 + 10x + 5)=(1 + sqrt(4(4x^2 + 10x + 5) +5))/2
=(1+sqrt(16x^2+40x+20+5))/2=(1+sqrt(16x^2+40x+25))/2=(1+(4x+5))/2=(4x+6)/2=2x+3}}} .
So, {{{highlight(g(x)=2x+3)}}} is a solution.
 
NOTES:
1) There was no problem with {{{f^-1(x)=(1 + sqrt(4x+5))/2}}}  being defined only for {{{x>=-5/4}}} ,
because we are going to apply it to
{{{4x^2 + 10x + 5=(2x+5/2)^2-5/4>=-5/4}}} .
2)
Would we get another solution out of {{{f^-1(x)=(1-sqrt(4x+5))/2}}} ?
Let's check.
In that case, we would have
{{{f^-1(4x^2 + 10x + 5)=(1-sqrt(4(4x^2 + 10x + 5) +5))/2
=(1-sqrt(16x^2+40x+20+5))/2=(1-sqrt(16x^2+40x+25))/2=(1-(4x+5))/2=(-4x-4)/2=-2x-2}}} .
If {{{f(-2x-2)=4x^2 + 10x + 5}}} ,
{{{f^-1(4x^2 + 10x + 5)=-2x-2}}} could  be another {{{g(x)}}} .
Let's see.
{{{f(-2x-2)=(-2x-2)^2-(-2x-2)-1=4x^2+4x+1+2x+2-1=4x^2+6x+1}}}
No. It does not work.