Solve Using the Quadratic Formula -x^2+2x=1

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

Step 1

Subtract $1$ from both sides of the equation.

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

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = - 1$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (- 1 \cdot - 1)}}{2 \cdot - 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot - 1}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot - 1$ .

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4 \cdot - 1}}{2 \cdot - 1}$

Multiply $4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4}}{2 \cdot - 1}$

Subtract $4$ from $4$ .

$x = \frac{- 2 \pm \sqrt{0}}{2 \cdot - 1}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{- 2 \pm \sqrt{0^{2}}}{2 \cdot - 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 2 \pm 0}{2 \cdot - 1}$

$- 2$ plus or minus $0$ is $- 2$ .

$x = \frac{- 2}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 2}{- 2}$

Divide $- 2$ by $- 2$ .

$x = 1$

Step 5

The final answer is the combination of both solutions.

$x = 1$ Double roots