Solve x=left(x-2right)^2-1

$$x = { \left(x-2 \right) }^{ 2 } -1$$

Solve for x

$$x = \frac{\sqrt{13} + 5}{2} \approx 4.302775638$$

$$x=\frac{5-\sqrt{13}}{2}\approx 0.697224362$$

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}.

$$x=x^{2}-4x+4-1$$

Subtract 1 from 4 to get 3.

$$x=x^{2}-4x+3$$

Subtract x^{2} from both sides.

$$x-x^{2}=-4x+3$$

Add 4x to both sides.

$$x-x^{2}+4x=3$$

Combine x and 4x to get 5x.

$$5x-x^{2}=3$$

Subtract 3 from both sides.

$$5x-x^{2}-3=0$$

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.

$$-x^{2}+5x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\left(-3\right)}}{2\left(-1\right)}$$

Multiply 4 times -3.

$$x=\frac{-5±\sqrt{25-12}}{2\left(-1\right)}$$

Add 25 to -12.

$$x=\frac{-5±\sqrt{13}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±\sqrt{13}}{-2}$$

Now solve the equation x=\frac{-5±\sqrt{13}}{-2} when ± is plus. Add -5 to \sqrt{13}.

$$x=\frac{\sqrt{13}-5}{-2}$$

Divide -5+\sqrt{13} by -2.

$$x=\frac{5-\sqrt{13}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{13}}{-2} when ± is minus. Subtract \sqrt{13} from -5.

$$x=\frac{-\sqrt{13}-5}{-2}$$

Divide -5-\sqrt{13} by -2.

$$x=\frac{\sqrt{13}+5}{2}$$

The equation is now solved.

$$x=\frac{5-\sqrt{13}}{2} x=\frac{\sqrt{13}+5}{2}$$

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}.

$$x=x^{2}-4x+4-1$$

Subtract 1 from 4 to get 3.

$$x=x^{2}-4x+3$$

Subtract x^{2} from both sides.

$$x-x^{2}=-4x+3$$

Add 4x to both sides.

$$x-x^{2}+4x=3$$

Combine x and 4x to get 5x.

$$5x-x^{2}=3$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+5x=3$$

Divide both sides by -1.

$$\frac{-x^{2}+5x}{-1}=\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{5}{-1}x=\frac{3}{-1}$$

Divide 5 by -1.

$$x^{2}-5x=\frac{3}{-1}$$

Divide 3 by -1.

$$x^{2}-5x=-3$$

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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-3+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-3+\frac{25}{4}$$

Add -3 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{13}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{13}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{13}}{2} x-\frac{5}{2}=-\frac{\sqrt{13}}{2}$$

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

$$x=\frac{\sqrt{13}+5}{2} x=\frac{5-\sqrt{13}}{2}$$