Solve 2x^2+4x-8=

$$2 { x }^{ 2 } +4x-8=$$

Factor

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

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$2x^{2}+4x-8=0$$

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.

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

Square 4.

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

Multiply -4 times 2.

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

Multiply -8 times -8.

$$x=\frac{-4±\sqrt{16+64}}{2\times 2}$$

Add 16 to 64.

$$x=\frac{-4±\sqrt{80}}{2\times 2}$$

Take the square root of 80.

$$x=\frac{-4±4\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±4\sqrt{5}}{4}$$

Now solve the equation x=\frac{-4±4\sqrt{5}}{4} when ± is plus. Add -4 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-4}{4}$$

Divide -4+4\sqrt{5} by 4.

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

Now solve the equation x=\frac{-4±4\sqrt{5}}{4} when ± is minus. Subtract 4\sqrt{5} from -4.

$$x=\frac{-4\sqrt{5}-4}{4}$$

Divide -4-4\sqrt{5} by 4.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1+\sqrt{5} for x_{1} and -1-\sqrt{5} for x_{2}.

$$2x^{2}+4x-8=2\left(x-\left(\sqrt{5}-1\right)\right)\left(x-\left(-\sqrt{5}-1\right)\right)$$

Evaluate

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