Solve 0.5x^2=x+0.5

$$0.5 { x }^{ 2 } =x+0.5$$

Solve for x

$$x=\sqrt{2}+1\approx 2.414213562$$

$$x=1-\sqrt{2}\approx -0.414213562$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$0.5x^{2}-x=0.5$$

Subtract 0.5 from both sides.

$$0.5x^{2}-x-0.5=0$$

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.

$$\frac{1}{2}x^{2}-x-0.5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -1 for b, and -0.5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{1}{2}\left(-0.5\right)}}{2\times \frac{1}{2}}$$

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

$$x=\frac{-\left(-1\right)±\sqrt{1-2\left(-0.5\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -0.5.

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

Add 1 to 1.

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

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{2}}{2\times \frac{1}{2}}$$

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

$$x=\frac{1±\sqrt{2}}{1}$$

Now solve the equation x=\frac{1±\sqrt{2}}{1} when ± is plus. Add 1 to \sqrt{2}.

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

Divide 1+\sqrt{2} by 1.

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

Now solve the equation x=\frac{1±\sqrt{2}}{1} when ± is minus. Subtract \sqrt{2} from 1.

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

Divide 1-\sqrt{2} by 1.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x from both sides.

$$0.5x^{2}-x=0.5$$

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.

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

Multiply both sides by 2.

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

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

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

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.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

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