Solve 0=x^2-4x-2

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

Solve for x

$$x=\sqrt{6}+2\approx 4.449489743$$

$$x=2-\sqrt{6}\approx -0.449489743$$

Steps Using the Quadratic Formula

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

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

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

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

Square -4.

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

Multiply -4 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+8}}{2}$$

Add 16 to 8.

$$x=\frac{-\left(-4\right)±\sqrt{24}}{2}$$

Take the square root of 24.

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

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{6}}{2}$$

Now solve the equation x=\frac{4±2\sqrt{6}}{2} when ± is plus. Add 4 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}+4}{2}$$

Divide 4+2\sqrt{6} by 2.

$$x=\sqrt{6}+2$$

Now solve the equation x=\frac{4±2\sqrt{6}}{2} when ± is minus. Subtract 2\sqrt{6} from 4.

$$x=\frac{4-2\sqrt{6}}{2}$$

Divide 4-2\sqrt{6} by 2.

$$x=2-\sqrt{6}$$

The equation is now solved.

$$x=\sqrt{6}+2 x=2-\sqrt{6}$$

Steps for Completing the Square

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

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

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

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add 2 to 4.

$$x^{2}-4x+4=6$$

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{6}$$

Simplify.

$$x-2=\sqrt{6} x-2=-\sqrt{6}$$

Add 2 to both sides of the equation.

$$x=\sqrt{6}+2 x=2-\sqrt{6}$$