Solve 0=-4x^2+4x+1

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

Solve for x

$$x = \frac{\sqrt{2} + 1}{2} \approx 1.207106781$$

$$x=\frac{1-\sqrt{2}}{2}\approx -0.207106781$$

Steps Using the Quadratic Formula

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

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

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-4\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-4±\sqrt{16+16}}{2\left(-4\right)}$$

Add 16 to 16.

$$x=\frac{-4±\sqrt{32}}{2\left(-4\right)}$$

Take the square root of 32.

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

Multiply 2 times -4.

$$x=\frac{-4±4\sqrt{2}}{-8}$$

Now solve the equation x=\frac{-4±4\sqrt{2}}{-8} when ± is plus. Add -4 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}-4}{-8}$$

Divide -4+4\sqrt{2} by -8.

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

Now solve the equation x=\frac{-4±4\sqrt{2}}{-8} when ± is minus. Subtract 4\sqrt{2} from -4.

$$x=\frac{-4\sqrt{2}-4}{-8}$$

Divide -4-4\sqrt{2} by -8.

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

The equation is now solved.

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

Steps for Completing the Square

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

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Divide 4 by -4.

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

Divide -1 by -4.

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

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.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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