Solve for x -x^2+x=0

$- x^{2} + x = 0$

Factor $- x$ out of $- x^{2} + x$ .

Factor $- x$ out of $- x^{2}$ .

$- x \cdot x + x = 0$

Rewrite $x$ as $1 x$ .

$- x \cdot x + 1 x = 0$

Factor $- x$ out of $1 x$ .

$- x \cdot x - x \cdot - 1 = 0$

Factor $- x$ out of $- x (x) - x (- 1)$ .

$- x (x - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $- x (x - 1) = 0$ true.

$x = 0, 1$