Roots to an equation, $\frac{3-2x}{x-1}-\sqrt[4]{x^3}$ 
Find the number of roots to $\frac{3-2x}{x-1}-\sqrt[4]{x^3}=0$.

My method:
$\frac{3-2x}{x-1}-\sqrt[4]{x}$, we multiply $\sqrt[4]{x}$ and $x-1$ to both sides, getting $3\sqrt[4]{x}-2x\sqrt[4]{x}-x(x-1)=0$.  We let $\sqrt[4]{x}=k$.  Then we have $3k-2xk-x^2+x=0$.  Finding the discriminant ($b^2-4ac$), getting $4k^2+8k+1$, evidently greater than $0$, which means there are 2 roots.
Plotting this on Desmos gives 1 root only. How am I wrong?

Since $k$ is a function of $x$, the coefficients of your quadratric equation are functions of $x$.

The quadratic formula is useless in that context, since that would be solving for $x$ in terms of $x$.

Here's how I would approach the problem . . .

Let $f(x)={\displaystyle{{\small{\frac{3-2x}{x-1}}}-\sqrt[4]{x^3}}}$.

You want to count the number of real zeros of $f$.

Rather than trying to solve the equation $f(x) = 0$, instead, analyze the behavior of $f$.

The domain of $f$ is $[0,1)\cup(1,\infty)$.

On the interval $[0,1)$, we have
$$\frac{3-2x}{x-1}<0$$
$$-\sqrt[4]{x^3} \le 0$$
hence $f(x) < 0$.

Thus, $f$ has no real zeros in the interval $[0,1)$.

On the interval $(1,\infty)$, the functions
$$\frac{3-2x}{x-1}$$
$$-\sqrt[4]{x^3}$$
are both strictly decreasing, hence $f$ is strictly decreasing.

It follows that $f$ has at most one real root.

Letting $x$ approach $1$ from the right, we get
\begin{align*}
\lim_{x\to 1^{+}} f(x)
&=\;\lim_{x\to 1^{+}} \frac{3-2x}{x-1}-\sqrt[4]{x^3}\\[4pt]
&=\;\lim_{x\to 1^{+}} \frac{3-2x}{x-1}-\lim_{x\to 1^{+}}\sqrt[4]{x^3}\\[4pt]
&=\;\infty-1\\[4pt]
&=\;\infty\\[4pt]
\end{align*}
Letting $x$ approach $\infty$, we get
\begin{align*}
\lim_{x\to \infty} f(x)
&=\;\lim_{x\to \infty} \frac{3-2x}{x-1}-\sqrt[4]{x^3}\\[4pt]
&=\;\lim_{x\to \infty} \frac{3-2x}{x-1}-\lim_{x\to \infty}\sqrt[4]{x^3}\\[4pt]
&=\;-2-\infty\\[4pt]
&=\;-\infty\\[4pt]
\end{align*}
Since $f$ is continuous on $(1,\infty)$, it follows that $f$ has at least one real root.

Therefore $f$ has exactly one real root.
