#### Answer

$x\in(0,1)$

#### Work Step by Step

Think of the equation as $f(x) \lt 0,\quad f(x)=x^{2}-x=x(x-1)$
A geometric interpretation is to find the values of $x$ for which the graph of $f(x)$ is below the x-axis.
The graph of $f$ is a parabola that opens up, because the leading coefficient is $+1$, a positive number.
$f(x)=x(x-1)$
so it has zeros (crosses the x-axis) at $x=0$ and $x=1$.
Since it opens up, the graph will be below the x-axis in the interval between the two zeros, $x\in(0,1)$.
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Solving algebraically, we analyze
$x(x-1) \lt 0$
asking "when is the product of two numbers on the LHS negative?"
Setting up a table where we test the signs of the product:
$\left[\begin{array}{ccccccc}
& -\infty & & 0 & & 1 & & \infty\\
\text{test point } t & & -1 & & 0.5 & & 2 & \\
f(t) & & +2 & & -0.25 & & +2 & \\
\text{sign of }f & & + & ( & - & ) & + &
\end{array}\right]$
We see that the inequality is true for $x\in(0,1)$