Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-x^{2}+2x-1=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx-1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=2$$ $$ab=-\left(-1\right)=1$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. The only such pair is the system solution.
$$a=1$$ $$b=1$$
Rewrite $-x^{2}+2x-1$ as $\left(-x^{2}+x\right)+\left(x-1\right)$.
$$\left(-x^{2}+x\right)+\left(x-1\right)$$
Factor out $-x$ in $-x^{2}+x$.
$$-x\left(x-1\right)+x-1$$
Factor out common term $x-1$ by using distributive property.
$$\left(x-1\right)\left(-x+1\right)$$
To find equation solutions, solve $x-1=0$ and $-x+1=0$.
$$x=1$$ $$x=1$$
Steps Using the Quadratic Formula
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$-x^{2}+2x+2=3$$
Subtract $3$ from both sides of the equation.
$$-x^{2}+2x+2-3=3-3$$
Subtracting $3$ from itself leaves $0$.
$$-x^{2}+2x+2-3=0$$
Subtract $3$ from $2$.
$$-x^{2}+2x-1=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $2$ for $b$, and $-1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.
$$-x^{2}+2x+2=3$$
Subtract $2$ from both sides of the equation.
$$-x^{2}+2x+2-2=3-2$$
Subtracting $2$ from itself leaves $0$.
$$-x^{2}+2x=3-2$$
Subtract $2$ from $3$.
$$-x^{2}+2x=1$$
Divide both sides by $-1$.
$$\frac{-x^{2}+2x}{-1}=\frac{1}{-1}$$
Dividing by $-1$ undoes the multiplication by $-1$.
$$x^{2}+\frac{2}{-1}x=\frac{1}{-1}$$
Divide $2$ by $-1$.
$$x^{2}-2x=\frac{1}{-1}$$
Divide $1$ by $-1$.
$$x^{2}-2x=-1$$
Divide $-2$, the coefficient of the $x$ term, by $2$ to get $-1$. Then add the square of $-1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}-2x+1=-1+1$$
Add $-1$ to $1$.
$$x^{2}-2x+1=0$$
Factor $x^{2}-2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x-1\right)^{2}=0$$
Take the square root of both sides of the equation.
$$\sqrt{\left(x-1\right)^{2}}=\sqrt{0}$$
Simplify.
$$x-1=0$$ $$x-1=0$$
Add $1$ to both sides of the equation.
$$x=1$$ $$x=1$$
The equation is now solved. Solutions are the same.
