Solve x^2+8x-7

$${ x }^{ 2 } +8x-7$$

Factor

$$\left(x-\left(-\sqrt{23}-4\right)\right)\left(x-\left(\sqrt{23}-4\right)\right)$$

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.

$$x^{2}+8x-7=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-7\right)}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-7\right)}}{2}$$

Multiply -4 times -7.

$$x=\frac{-8±\sqrt{64+28}}{2}$$

Add 64 to 28.

$$x=\frac{-8±\sqrt{92}}{2}$$

Take the square root of 92.

$$x=\frac{-8±2\sqrt{23}}{2}$$

Now solve the equation x=\frac{-8±2\sqrt{23}}{2} when ± is plus. Add -8 to 2\sqrt{23}.

$$x=\frac{2\sqrt{23}-8}{2}$$

Divide -8+2\sqrt{23} by 2.

$$x=\sqrt{23}-4$$

Now solve the equation x=\frac{-8±2\sqrt{23}}{2} when ± is minus. Subtract 2\sqrt{23} from -8.

$$x=\frac{-2\sqrt{23}-8}{2}$$

Divide -8-2\sqrt{23} by 2.

$$x=-\sqrt{23}-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4+\sqrt{23} for x_{1} and -4-\sqrt{23} for x_{2}.

$$x^{2}+8x-7=\left(x-\left(\sqrt{23}-4\right)\right)\left(x-\left(-\sqrt{23}-4\right)\right)$$

Evaluate

$$x^{2}+8x-7$$