Solve for x -x^2+2x=1

$- x^{2} + 2 x = 1$

Subtract $1$ from both sides of the equation.

$- x^{2} + 2 x - 1 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 2 x - 1$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 2 x - 1 = 0$

Factor $- 1$ out of $2 x$ .

$- (x^{2}) - (- 2 x) - 1 = 0$

Rewrite $- 1$ as $- 1 (1)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot 1 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 2 x)$ .

$- (x^{2} - 2 x) - 1 \cdot 1 = 0$

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (1)$ .

$- (x^{2} - 2 x + 1) = 0$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$- (x^{2} - 2 x + 1^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$- (x^{2} - 2 \cdot x \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

$- {(x - 1)}^{2} = 0$

Divide each term in $- {(x - 1)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(x - 1)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 1)}^{2}}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(x - 1)}^{2}}{1} = \frac{0}{- 1}$

Divide ${(x - 1)}^{2}$ by $1$ .

${(x - 1)}^{2} = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

${(x - 1)}^{2} = 0$

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$