Solve 0=1/2x^2-x-2

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

Solve for x

$$x=\sqrt{5}+1\approx 3.236067977$$

$$x=1-\sqrt{5}\approx -1.236067977$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

Multiply -2 times -2.

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

Add 1 to 4.

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

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{5}}{2\times \frac{1}{2}}$$

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

$$x=\frac{1±\sqrt{5}}{1}$$

Now solve the equation x=\frac{1±\sqrt{5}}{1} when ± is plus. Add 1 to \sqrt{5}.

$$x=\frac{\sqrt{5}+1}{1}$$

Divide 1+\sqrt{5} by 1.

$$x=\sqrt{5}+1$$

Now solve the equation x=\frac{1±\sqrt{5}}{1} when ± is minus. Subtract \sqrt{5} from 1.

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

Divide 1-\sqrt{5} by 1.

$$x=1-\sqrt{5}$$

The equation is now solved.

$$x=\sqrt{5}+1 x=1-\sqrt{5}$$

Steps for Completing the Square

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

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

Add 2 to both sides. Anything plus zero gives itself.

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

Multiply both sides by 2.

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

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

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

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

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

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

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

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.

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

Add 4 to 1.

$$x^{2}-2x+1=5$$

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{5}$$

Simplify.

$$x-1=\sqrt{5} x-1=-\sqrt{5}$$

Add 1 to both sides of the equation.

$$x=\sqrt{5}+1 x=1-\sqrt{5}$$