Find $a$ so that $f$ has a specific local extreme point We are given a function $f : \mathbb{R} \rightarrow \mathbb{R}$ with:

$$f(x) = \frac{x^2 - ax}{\sqrt{x^2 + 1}}, a \in \mathbb{R}$$ 

Find $a$ so that the function $f$ has a point of local extreme situated at a distance of $2$ from the $Oy$ axis.

The derivative of a function in a local extreme point is $0$. Calculating the derivative of $f$ and then equalizing with $0$, I got $x^3 + 2x - a = 0$.

At this point I got stuck.

Thank you in advance!
let $$h(x)=x^3+2x-a$$

$$h(2)=0 \rightarrow a=13$$

$$h(-2)=0 \rightarrow a=-13$$

When $-13 \lt a \lt 13$,  $h(2) \gt 0$ and $h(-2) \lt 0$ 
Thus by intermediate value theorem, there is a $c \in (-2, 2)$ such that $h(c)=0$. We can conclude there is an critical point within (-2, 2).
let $$h(x)=x^3+2x-a$$

$$h(2)=0 \rightarrow a=13$$

$$h(-2)=0 \rightarrow a=-13$$

When $-13 \lt a \lt 13$,  $h(2) \gt 0$ and $h(-2) \lt 0$ 
Thus by intermediate value theorem, there is a $c \in (-2, 2)$ such that $h(c)=0$. We can conclude there is an critical point within (-2, 2).