Solve x=(x+2)^2-1

$$x = ( x + 2 ) ^ { 2 } - 1$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{3}i-3}{2}\approx -1.5-0.866025404i$$

$$x=\frac{-3+\sqrt{3}i}{2}\approx -1.5+0.866025404i$$

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.

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

Subtract 4x from both sides.

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

Combine x and -4x to get -3x.

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

Subtract 3 from both sides.

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

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.

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

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

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\left(-3\right)}}{2\left(-1\right)}$$

Multiply 4 times -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-12}}{2\left(-1\right)}$$

Add 9 to -12.

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

Take the square root of -3.

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

The opposite of -3 is 3.

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

Multiply 2 times -1.

$$x=\frac{3±\sqrt{3}i}{-2}$$

Now solve the equation x=\frac{3±\sqrt{3}i}{-2} when ± is plus. Add 3 to i\sqrt{3}.

$$x=\frac{3+\sqrt{3}i}{-2}$$

Divide 3+i\sqrt{3} by -2.

$$x=\frac{-\sqrt{3}i-3}{2}$$

Now solve the equation x=\frac{3±\sqrt{3}i}{-2} when ± is minus. Subtract i\sqrt{3} from 3.

$$x=\frac{-\sqrt{3}i+3}{-2}$$

Divide 3-i\sqrt{3} by -2.

$$x=\frac{-3+\sqrt{3}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{3}i-3}{2} x=\frac{-3+\sqrt{3}i}{2}$$

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.

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

Subtract 4x from both sides.

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

Combine x and -4x to get -3x.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{3}{-1}\right)x=\frac{3}{-1}$$

Divide -3 by -1.

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

Divide 3 by -1.

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

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-3+\frac{9}{4}$$

Add -3 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=-\frac{3}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{3}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{3}i}{2} x+\frac{3}{2}=-\frac{\sqrt{3}i}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{-3+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i-3}{2}$$