Solve -1/4=x^2-x

$$- \frac { 1 } { 4 } = x ^ { 2 } - x$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-x=-\frac{1}{4}$$

Add \frac{1}{4} to both sides.

$$x^{2}-x+\frac{1}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and \frac{1}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{1}{4}}}{2}$$

Multiply -4 times \frac{1}{4}.

$$x=\frac{-\left(-1\right)±\sqrt{1-1}}{2}$$

Add 1 to -1.

$$x=\frac{-\left(-1\right)±\sqrt{0}}{2}$$

Take the square root of 0.

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

The opposite of -1 is 1.

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

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-x=-\frac{1}{4}$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{1}{4}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{-1+1}{4}$$

Add -\frac{1}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=0$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

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

The equation is now solved. Solutions are the same.

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