Solve 0=-1/2(x-1/2)^2+25/8

$$0 = - \frac { 1 } { 2 } ( x - \frac { 1 } { 2 } ) ^ { 2 } + \frac { 25 } { 8 }$$

Solve for x

$$x=3$$

$$x=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{2}\right)^{2}.

$$0=-\frac{1}{2}\left(x^{2}-x+\frac{1}{4}\right)+\frac{25}{8}$$

Use the distributive property to multiply -\frac{1}{2} by x^{2}-x+\frac{1}{4}.

$$0=-\frac{1}{2}x^{2}+\frac{1}{2}x-\frac{1}{8}+\frac{25}{8}$$

Add -\frac{1}{8} and \frac{25}{8} to get 3.

$$0=-\frac{1}{2}x^{2}+\frac{1}{2}x+3$$

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

$$-\frac{1}{2}x^{2}+\frac{1}{2}x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{2} for a, \frac{1}{2} for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\left(-\frac{1}{2}\right)\times 3}}{2\left(-\frac{1}{2}\right)}$$

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

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\left(-\frac{1}{2}\right)\times 3}}{2\left(-\frac{1}{2}\right)}$$

Multiply -4 times -\frac{1}{2}.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+2\times 3}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times 3.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+6}}{2\left(-\frac{1}{2}\right)}$$

Add \frac{1}{4} to 6.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{25}{4}}}{2\left(-\frac{1}{2}\right)}$$

Take the square root of \frac{25}{4}.

$$x=\frac{-\frac{1}{2}±\frac{5}{2}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{1}{2}.

$$x=\frac{-\frac{1}{2}±\frac{5}{2}}{-1}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{5}{2}}{-1} when ± is plus. Add -\frac{1}{2} to \frac{5}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 2 by -1.

$$x=-2$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{5}{2}}{-1} when ± is minus. Subtract \frac{5}{2} from -\frac{1}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{3}{-1}$$

Divide -3 by -1.

$$x=3$$

The equation is now solved.

$$x=-2 x=3$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{2}\right)^{2}.

$$0=-\frac{1}{2}\left(x^{2}-x+\frac{1}{4}\right)+\frac{25}{8}$$

Use the distributive property to multiply -\frac{1}{2} by x^{2}-x+\frac{1}{4}.

$$0=-\frac{1}{2}x^{2}+\frac{1}{2}x-\frac{1}{8}+\frac{25}{8}$$

Add -\frac{1}{8} and \frac{25}{8} to get 3.

$$0=-\frac{1}{2}x^{2}+\frac{1}{2}x+3$$

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

$$-\frac{1}{2}x^{2}+\frac{1}{2}x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$-\frac{1}{2}x^{2}+\frac{1}{2}x=-3$$

Multiply both sides by -2.

$$\frac{-\frac{1}{2}x^{2}+\frac{1}{2}x}{-\frac{1}{2}}=-\frac{3}{-\frac{1}{2}}$$

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

$$x^{2}+\frac{\frac{1}{2}}{-\frac{1}{2}}x=-\frac{3}{-\frac{1}{2}}$$

Divide \frac{1}{2} by -\frac{1}{2} by multiplying \frac{1}{2} by the reciprocal of -\frac{1}{2}.

$$x^{2}-x=-\frac{3}{-\frac{1}{2}}$$

Divide -3 by -\frac{1}{2} by multiplying -3 by the reciprocal of -\frac{1}{2}.

$$x^{2}-x=6$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=6+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{25}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}$$

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

$$x=3 x=-2$$