Solve x^2+8x-4

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

Factor

$$\left(x-\left(-2\sqrt{5}-4\right)\right)\left(x-\left(2\sqrt{5}-4\right)\right)$$

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.

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

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.

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

Square 8.

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

Multiply -4 times -4.

$$x=\frac{-8±\sqrt{64+16}}{2}$$

Add 64 to 16.

$$x=\frac{-8±\sqrt{80}}{2}$$

Take the square root of 80.

$$x=\frac{-8±4\sqrt{5}}{2}$$

Now solve the equation x=\frac{-8±4\sqrt{5}}{2} when ± is plus. Add -8 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-8}{2}$$

Divide -8+4\sqrt{5} by 2.

$$x=2\sqrt{5}-4$$

Now solve the equation x=\frac{-8±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from -8.

$$x=\frac{-4\sqrt{5}-8}{2}$$

Divide -8-4\sqrt{5} by 2.

$$x=-2\sqrt{5}-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4+2\sqrt{5} for x_{1} and -4-2\sqrt{5} for x_{2}.

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

Evaluate

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