Solve (x-2)^2-4x+2=0

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

Solve for x

$$x=\sqrt{10}+4\approx 7.16227766$$

$$x=4-\sqrt{10}\approx 0.83772234$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

Combine -4x and -4x to get -8x.

$$x^{2}-8x+4+2=0$$

Add 4 and 2 to get 6.

$$x^{2}-8x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 6}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 6}}{2}$$

Multiply -4 times 6.

$$x=\frac{-\left(-8\right)±\sqrt{64-24}}{2}$$

Add 64 to -24.

$$x=\frac{-\left(-8\right)±\sqrt{40}}{2}$$

Take the square root of 40.

$$x=\frac{-\left(-8\right)±2\sqrt{10}}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{10}}{2}$$

Now solve the equation x=\frac{8±2\sqrt{10}}{2} when ± is plus. Add 8 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}+8}{2}$$

Divide 8+2\sqrt{10} by 2.

$$x=\sqrt{10}+4$$

Now solve the equation x=\frac{8±2\sqrt{10}}{2} when ± is minus. Subtract 2\sqrt{10} from 8.

$$x=\frac{8-2\sqrt{10}}{2}$$

Divide 8-2\sqrt{10} by 2.

$$x=4-\sqrt{10}$$

The equation is now solved.

$$x=\sqrt{10}+4 x=4-\sqrt{10}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

Combine -4x and -4x to get -8x.

$$x^{2}-8x+4+2=0$$

Add 4 and 2 to get 6.

$$x^{2}-8x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-8x=-6$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-6+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-6+16$$

Add -6 to 16.

$$x^{2}-8x+16=10$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=10$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{10}$$

Simplify.

$$x-4=\sqrt{10} x-4=-\sqrt{10}$$

Add 4 to both sides of the equation.

$$x=\sqrt{10}+4 x=4-\sqrt{10}$$