Solve x^2+6x-8

$${ x }^{ 2 } +6x-8$$

Factor

$$\left(x-\left(-\sqrt{17}-3\right)\right)\left(x-\left(\sqrt{17}-3\right)\right)$$

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.

$$x^{2}+6x-8=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-8\right)}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-8\right)}}{2}$$

Multiply -4 times -8.

$$x=\frac{-6±\sqrt{36+32}}{2}$$

Add 36 to 32.

$$x=\frac{-6±\sqrt{68}}{2}$$

Take the square root of 68.

$$x=\frac{-6±2\sqrt{17}}{2}$$

Now solve the equation x=\frac{-6±2\sqrt{17}}{2} when ± is plus. Add -6 to 2\sqrt{17}.

$$x=\frac{2\sqrt{17}-6}{2}$$

Divide -6+2\sqrt{17} by 2.

$$x=\sqrt{17}-3$$

Now solve the equation x=\frac{-6±2\sqrt{17}}{2} when ± is minus. Subtract 2\sqrt{17} from -6.

$$x=\frac{-2\sqrt{17}-6}{2}$$

Divide -6-2\sqrt{17} by 2.

$$x=-\sqrt{17}-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3+\sqrt{17} for x_{1} and -3-\sqrt{17} for x_{2}.

$$x^{2}+6x-8=\left(x-\left(\sqrt{17}-3\right)\right)\left(x-\left(-\sqrt{17}-3\right)\right)$$

Evaluate

$$x^{2}+6x-8$$