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.
$$\frac{1}{2}x^{2}-x+\frac{1}{2}=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $\frac{1}{2}$ for $a$, $-1$ for $b$, and $\frac{1}{2}$ 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 $x^{2}+bx=c$.
$$\frac{1}{2}x^{2}-x+\frac{1}{2}=0$$
Subtract $\frac{1}{2}$ from both sides of the equation.
Divide $-1$ by $\frac{1}{2}$ by multiplying $-1$ by the reciprocal of $\frac{1}{2}$.
$$x^{2}-2x=-\frac{\frac{1}{2}}{\frac{1}{2}}$$
Divide $-\frac{1}{2}$ by $\frac{1}{2}$ by multiplying $-\frac{1}{2}$ by the reciprocal of $\frac{1}{2}$.
$$x^{2}-2x=-1$$
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.
