(To state my solution more rigorously and clearly, I put my comment into an answer.)
First, it is easy to calculate:
$$
f'(x) = e^x (x^2+5x+1) - 2x -2 \\
f''(x) = (x+1)(x+6)e^x-2 \\
f'''(x)=e^x(x^2+9x+13)
$$
Second, we will check how many real roots $f''(x)=0$ has:
Case $-\infty<x \leq -1$:
Calculate $f'''(x)=0$, we get the stationary points of $f''(x)$ are at
$$ x_{1} = \frac{-9 + \sqrt{29}}{2}, ~ x_{2} = \frac{-9 - \sqrt{29}}{2}$$
Then evaluate: $f''(x_1)\approx -2.55543$, $f''(x_2)\approx -1.99445$. Besides,
$$ \lim_{x \rightarrow -\infty} f''(x) = -2$$
and $f''(-1)=-2$. Thus,
$$\max_{-\infty < x \leq -1} f''(x)=f''(x_2)\approx -1.99445 < 0$$
That is, in this case, there's no real root for $f''(x)=0$.
Case $x>-1$:
In this case, it is easy to check $f''(x)$ is monotone increasing and is able to go larger than 0. Thus, there's exactly one real root for $f''(x)=0, x>-1$.
Conclusion: $f''(x)=0$ has exactly one real root.
Finally, we will utilize Rolle's theorem:
(Given proper continuity and differentiability.) If $f'(x)=0$ has exactly $n$ real roots, then $f(x)=0$ at most has $n+1$ real roots.
Why? If we can find $n+2$ real roots for $f(x)=0$, say
$$x_1<x_2<\cdots<x_{n+2}$$
according to Rolle's theorem, there will exist $c_1 \in (x_1,x_2)$, $c_2 \in (x_2,x_3)$, ..., $c_{n+1} \in (x_{n+1},x_{n+2})$, such that
$$f'(c_1) = f'(c_2) = \cdots = f'(c_{n+1}) = 0$$
It will contradict "$f'(x)=0$ has exactly $n$ real roots".
Now, we can see because $f''(x)=0$ has exactly one real root, $f'(x)=0$ will at most have two real roots, and then $f(x)=0$ will at most have three real roots. And you have found the three roots for $f(x)=0$, so there're no more roots.
Note: Be careful of the "at most" statements above. It does
NOT mean the equation must have such many roots.
Suppose that $f'(a)=f'(b)=0$ with $a<b$ and that $f'\not = 0$ in $]a,b[$. If $f$ has two distinct roots $x$ and $y$ in $]a,b[$, then, by Rolle's theorem (since $f(x)=f(y)=0$), there is $c$ in $[x,y]$ such that $f'(c)=0$. Contradiction.
Best Answer
Yes, a root is basically the same as a solution. We often use the work "root" when we talk about polynomials.
Now, this doesn't mean that the words are interchangeable. We can say
We can't say
Solutions are always to equations. Roots are for/of polynomials (functions).
You write that the equation $$ \frac{1}{\ln(x)} - \frac{1}{x-1} $$ has a solution, but what you have written isn't an equation because you don't have an equal sign. Instead you probably want to say that the equation $$ \frac{1}{\ln(x)} - \frac{1}{x-1} =0 $$ has a solution ...
Example: Let $f(x) = x^2 -3x + 2$. Then $f$ is a polynomial. We can say the following