Solve x^2-18x+9

$${ x }^{ 2 } -18x+9$$

Factor

$$\left(x-\left(9-6\sqrt{2}\right)\right)\left(x-\left(6\sqrt{2}+9\right)\right)$$

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.

$$x^{2}-18x+9=0$$

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.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 9}}{2}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 9}}{2}$$

Multiply -4 times 9.

$$x=\frac{-\left(-18\right)±\sqrt{324-36}}{2}$$

Add 324 to -36.

$$x=\frac{-\left(-18\right)±\sqrt{288}}{2}$$

Take the square root of 288.

$$x=\frac{-\left(-18\right)±12\sqrt{2}}{2}$$

The opposite of -18 is 18.

$$x=\frac{18±12\sqrt{2}}{2}$$

Now solve the equation x=\frac{18±12\sqrt{2}}{2} when ± is plus. Add 18 to 12\sqrt{2}.

$$x=\frac{12\sqrt{2}+18}{2}$$

Divide 18+12\sqrt{2} by 2.

$$x=6\sqrt{2}+9$$

Now solve the equation x=\frac{18±12\sqrt{2}}{2} when ± is minus. Subtract 12\sqrt{2} from 18.

$$x=\frac{18-12\sqrt{2}}{2}$$

Divide 18-12\sqrt{2} by 2.

$$x=9-6\sqrt{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 9+6\sqrt{2} for x_{1} and 9-6\sqrt{2} for x_{2}.

$$x^{2}-18x+9=\left(x-\left(6\sqrt{2}+9\right)\right)\left(x-\left(9-6\sqrt{2}\right)\right)$$

Evaluate

$$x^{2}-18x+9$$