Solve (x^2+2x-9)

$$( { x }^{ 2 } +2x-9)$$

Factor

$$\left(x-\left(-\sqrt{10}-1\right)\right)\left(x-\left(\sqrt{10}-1\right)\right)$$

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.

$$x^{2}+2x-9=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-9\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$x=\frac{-2±\sqrt{4+36}}{2}$$

Add 4 to 36.

$$x=\frac{-2±\sqrt{40}}{2}$$

Take the square root of 40.

$$x=\frac{-2±2\sqrt{10}}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{10}}{2} when ± is plus. Add -2 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-2}{2}$$

Divide -2+2\sqrt{10} by 2.

$$x=\sqrt{10}-1$$

Now solve the equation x=\frac{-2±2\sqrt{10}}{2} when ± is minus. Subtract 2\sqrt{10} from -2.

$$x=\frac{-2\sqrt{10}-2}{2}$$

Divide -2-2\sqrt{10} by 2.

$$x=-\sqrt{10}-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1+\sqrt{10} for x_{1} and -1-\sqrt{10} for x_{2}.

$$x^{2}+2x-9=\left(x-\left(\sqrt{10}-1\right)\right)\left(x-\left(-\sqrt{10}-1\right)\right)$$

Evaluate

$$x^{2}+2x-9$$