Solve x^2+2-4x

$$x ^ { 2 } + 2 - 4 x$$

Factor

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

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.

$$x^{2}-4x+2=0$$

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.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8}}{2}$$

Add 16 to -8.

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

Take the square root of 8.

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

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{2}}{2}$$

Now solve the equation x=\frac{4±2\sqrt{2}}{2} when ± is plus. Add 4 to 2\sqrt{2}.

$$x=\frac{2\sqrt{2}+4}{2}$$

Divide 4+2\sqrt{2} by 2.

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

Now solve the equation x=\frac{4±2\sqrt{2}}{2} when ± is minus. Subtract 2\sqrt{2} from 4.

$$x=\frac{4-2\sqrt{2}}{2}$$

Divide 4-2\sqrt{2} by 2.

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

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

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

Evaluate

$$x^{2}-4x+2$$