Show that $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{3x-4}{x^2+5}$ is injective I'm working on proving that $f: \Bbb{R} \rightarrow \Bbb{R}$ defined by $f(x) = \frac{3x-4}{x^2+5}$ is injective. However I'm stuck.
Assuming that $f(x_1) = f(x_2)$:
$$f(x_1) = f(x_2)$$
$$\frac{3x_1-4}{x_1^2+5} = \frac{3x_2-4}{x_2^2+5}$$
$$(3x_1-4)(x_2^2+5) = (3x_2-4)(x_1^2+5)$$
$$3x_1x_2^2 + 15x_1 - 4x_2^2 - 20 = 3x_2x_1^2 + 15x_2 - 4x_1^2 - 20$$
$$3x_1x_2^2 + 15x_1 - 4x_2^2 = 3x_2x_1^2 + 15x_2 - 4x_1^2$$
How do I get to $x_1 = x_2$ from here on out?
 A: Desmos showing the graph of this function is :
The horizontal line near the $y$ value $-1$(for example) intersect twice the graph ,concluding this function is not one-one! 
A: You can show the non-injectivity of the function without finding specific values $x_0 \neq x_1$ with $f(x_0)=f(x_1)$ as follows:


*

*$f(0) = -\frac{4}{5}$

*$\lim_{x\to \color{blue}{-\infty}}f(x)= 0$

*$\lim_{x\to \color{blue}{+\infty}}f(x)= 0$
Applying the intermediate-value property for continuous functions you immediately get the existence of $x_0 \in (\color{blue}{-\infty},0)$ and $x_1 \in (0,\color{blue}{+\infty})$ with $f(x_0) = f(x_1) = -\frac{1}{4}$. Done.
A: When a function is injective, you have that $f(x_1) = f(x_2)$ implies in $x_1 = x_2$ for all elements in the domain of the function you are working with.
Take $f(0)=f\left(\dfrac{-15}4\right)$ as @J.W.Tanner said in the comments. In that case, you have $f(x_1) = f(x_2)$. However, $x_1$ is not equal to $x_2$. 
A counterexample to the implication, since you have at least one element that makes it not happen.
$f(x_1) = f(x_2)$ does not imply in $x_1 = x_2$.
So, the function $f(x) = \frac{3x-4}{x^2+5}$ is not injective (one-to-one).
A: $$f(x) = \frac{3x-4}{x^2+5}$$
$$f(0)=-\dfrac 45 = f\left(-\dfrac{15}{4}\right)$$
You would have had an easier time solving 
\begin{align}
   \dfrac{3x-4}{x^2+5} &= k \\
   3x-4 &= kx^2 + 5k \\
   kx^2 - 3x +(5k+4) &= 0 \\
   x &= \dfrac{3 \pm \sqrt{9-16k-20k^2}}{2k}
\end{align}
So $x$ is double valued for all 
$$-\dfrac 25 - \dfrac{\sqrt{61}}{10} < k < -\dfrac 25 + \dfrac{\sqrt{61}}{10}$$
except it is single-valued at $k=0$
A: By your work we obtain $$3x_1x_2(x_1-x_2)-15(x_1-x_2)-4(x_1^2-x_2^2)=0$$ or
$$(x_1-x_2)(3x_1x_2-15-4(x_1+x_2))=0,$$ which gives also
$$3x_1x_2-15-4(x_1+x_2)=0$$ for $x_1\neq x_2,$ which says that our function is not injective. 
