Solve 2x^2-8x-2*6=0

$$2 x ^ { 2 } - 8 x - 2 \cdot 6 = 0$$

Solve for x

$$x=\sqrt{10}+2\approx 5.16227766$$

$$x=2-\sqrt{10}\approx -1.16227766$$

Steps Using the Quadratic Formula

Multiply 2 and 6 to get 12.

$$2x^{2}-8x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-12\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\left(-12\right)}}{2\times 2}$$

Multiply -8 times -12.

$$x=\frac{-\left(-8\right)±\sqrt{64+96}}{2\times 2}$$

Add 64 to 96.

$$x=\frac{-\left(-8\right)±\sqrt{160}}{2\times 2}$$

Take the square root of 160.

$$x=\frac{-\left(-8\right)±4\sqrt{10}}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{10}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±4\sqrt{10}}{4}$$

Now solve the equation x=\frac{8±4\sqrt{10}}{4} when ± is plus. Add 8 to 4\sqrt{10}.

$$x=\frac{4\sqrt{10}+8}{4}$$

Divide 8+4\sqrt{10} by 4.

$$x=\sqrt{10}+2$$

Now solve the equation x=\frac{8±4\sqrt{10}}{4} when ± is minus. Subtract 4\sqrt{10} from 8.

$$x=\frac{8-4\sqrt{10}}{4}$$

Divide 8-4\sqrt{10} by 4.

$$x=2-\sqrt{10}$$

The equation is now solved.

$$x=\sqrt{10}+2 x=2-\sqrt{10}$$

Steps for Completing the Square

Multiply 2 and 6 to get 12.

$$2x^{2}-8x-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$2x^{2}-8x=12$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=\frac{12}{2}$$

Divide -8 by 2.

$$x^{2}-4x=\frac{12}{2}$$

Divide 12 by 2.

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

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.

$$x^{2}-4x+\left(-2\right)^{2}=6+\left(-2\right)^{2}$$

Square -2.

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

Add 6 to 4.

$$x^{2}-4x+4=10$$

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}.

$$\left(x-2\right)^{2}=10$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{10}$$

Simplify.

$$x-2=\sqrt{10} x-2=-\sqrt{10}$$

Add 2 to both sides of the equation.

$$x=\sqrt{10}+2 x=2-\sqrt{10}$$