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.
$$2x^{2}-x+\frac{1}{8}=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-1$ for $b$, and $\frac{1}{8}$ 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 $x^{2}+bx=c$.
$$2x^{2}-x+\frac{1}{8}=0$$
Subtract $\frac{1}{8}$ from both sides of the equation.
$$2x^{2}-x+\frac{1}{8}-\frac{1}{8}=-\frac{1}{8}$$
Subtracting $\frac{1}{8}$ from itself leaves $0$.
$$2x^{2}-x=-\frac{1}{8}$$
Divide both sides by $2$.
$$\frac{2x^{2}-x}{2}=-\frac{\frac{1}{8}}{2}$$
Dividing by $2$ undoes the multiplication by $2$.
$$x^{2}-\frac{1}{2}x=-\frac{\frac{1}{8}}{2}$$
Divide $-\frac{1}{8}$ by $2$.
$$x^{2}-\frac{1}{2}x=-\frac{1}{16}$$
Divide $-\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{4}$. Then add the square of $-\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
Add $-\frac{1}{16}$ to $\frac{1}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
$$x^{2}-\frac{1}{2}x+\frac{1}{16}=0$$
Factor $x^{2}-\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x-\frac{1}{4}\right)^{2}=0$$
Take the square root of both sides of the equation.
