Solve 2x^2-18x+9

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

Factor

$$2\left(x-\frac{9-3\sqrt{7}}{2}\right)\left(x-\frac{3\sqrt{7}+9}{2}\right)$$

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.

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

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.

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

Square -18.

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

Multiply -4 times 2.

$$x=\frac{-\left(-18\right)±\sqrt{324-8\times 9}}{2\times 2}$$

Multiply -8 times 9.

$$x=\frac{-\left(-18\right)±\sqrt{324-72}}{2\times 2}$$

Add 324 to -72.

$$x=\frac{-\left(-18\right)±\sqrt{252}}{2\times 2}$$

Take the square root of 252.

$$x=\frac{-\left(-18\right)±6\sqrt{7}}{2\times 2}$$

The opposite of -18 is 18.

$$x=\frac{18±6\sqrt{7}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{18±6\sqrt{7}}{4}$$

Now solve the equation x=\frac{18±6\sqrt{7}}{4} when ± is plus. Add 18 to 6\sqrt{7}.

$$x=\frac{6\sqrt{7}+18}{4}$$

Divide 18+6\sqrt{7} by 4.

$$x=\frac{3\sqrt{7}+9}{2}$$

Now solve the equation x=\frac{18±6\sqrt{7}}{4} when ± is minus. Subtract 6\sqrt{7} from 18.

$$x=\frac{18-6\sqrt{7}}{4}$$

Divide 18-6\sqrt{7} by 4.

$$x=\frac{9-3\sqrt{7}}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9+3\sqrt{7}}{2} for x_{1} and \frac{9-3\sqrt{7}}{2} for x_{2}.

$$2x^{2}-18x+9=2\left(x-\frac{3\sqrt{7}+9}{2}\right)\left(x-\frac{9-3\sqrt{7}}{2}\right)$$

Evaluate

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