Solve (x-2)^2-x=2

$$( x - 2 ) ^ { 2 } - x = 2$$

Solve for x

$$x = \frac{\sqrt{17} + 5}{2} \approx 4.561552813$$

$$x=\frac{5-\sqrt{17}}{2}\approx 0.438447187$$

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}.

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

Combine -4x and -x to get -5x.

$$x^{2}-5x+4=2$$

Subtract 2 from both sides.

$$x^{2}-5x+4-2=0$$

Subtract 2 from 4 to get 2.

$$x^{2}-5x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8}}{2}$$

Add 25 to -8.

$$x=\frac{-\left(-5\right)±\sqrt{17}}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{17}}{2}$$

Now solve the equation x=\frac{5±\sqrt{17}}{2} when ± is plus. Add 5 to \sqrt{17}.

$$x=\frac{\sqrt{17}+5}{2}$$

Now solve the equation x=\frac{5±\sqrt{17}}{2} when ± is minus. Subtract \sqrt{17} from 5.

$$x=\frac{5-\sqrt{17}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+5}{2} x=\frac{5-\sqrt{17}}{2}$$

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}.

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

Combine -4x and -x to get -5x.

$$x^{2}-5x+4=2$$

Subtract 4 from both sides.

$$x^{2}-5x=2-4$$

Subtract 4 from 2 to get -2.

$$x^{2}-5x=-2$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-2+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-2+\frac{25}{4}$$

Add -2 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{17}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{17}}{2} x-\frac{5}{2}=-\frac{\sqrt{17}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{17}+5}{2} x=\frac{5-\sqrt{17}}{2}$$