Solve x^2+4x-8=

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

Factor

$$\left(x-\left(-2\sqrt{3}-2\right)\right)\left(x-\left(2\sqrt{3}-2\right)\right)$$

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.

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

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.

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

Square 4.

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

Multiply -4 times -8.

$$x=\frac{-4±\sqrt{16+32}}{2}$$

Add 16 to 32.

$$x=\frac{-4±\sqrt{48}}{2}$$

Take the square root of 48.

$$x=\frac{-4±4\sqrt{3}}{2}$$

Now solve the equation x=\frac{-4±4\sqrt{3}}{2} when ± is plus. Add -4 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}-4}{2}$$

Divide -4+4\sqrt{3} by 2.

$$x=2\sqrt{3}-2$$

Now solve the equation x=\frac{-4±4\sqrt{3}}{2} when ± is minus. Subtract 4\sqrt{3} from -4.

$$x=\frac{-4\sqrt{3}-4}{2}$$

Divide -4-4\sqrt{3} by 2.

$$x=-2\sqrt{3}-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2+2\sqrt{3} for x_{1} and -2-2\sqrt{3} for x_{2}.

$$x^{2}+4x-8=\left(x-\left(2\sqrt{3}-2\right)\right)\left(x-\left(-2\sqrt{3}-2\right)\right)$$

Evaluate

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