Solve x-left(2+xright)^2=-2

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

Solve for x

$$x=-2$$

$$x=-1$$

Steps Using Factoring By Grouping

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

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

To find the opposite of 4+4x+x^{2}, find the opposite of each term.

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

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

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

Add 2 to both sides.

$$-3x-4-x^{2}+2=0$$

Add -4 and 2 to get -2.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-\left(-2\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-2$$

Rewrite -x^{2}-3x-2 as \left(-x^{2}-x\right)+\left(-2x-2\right).

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

Factor out x in the first and 2 in the second group.

$$x\left(-x-1\right)+2\left(-x-1\right)$$

Factor out common term -x-1 by using distributive property.

$$\left(-x-1\right)\left(x+2\right)$$

To find equation solutions, solve -x-1=0 and x+2=0.

$$x=-1 x=-2$$

Steps Using the Quadratic Formula

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

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

To find the opposite of 4+4x+x^{2}, find the opposite of each term.

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

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

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

Add 2 to both sides.

$$-3x-4-x^{2}+2=0$$

Add -4 and 2 to get -2.

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

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.

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

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

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

Square -3.

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

Multiply -4 times -1.

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

Multiply 4 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2\left(-1\right)}$$

Add 9 to -8.

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

Take the square root of 1.

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

The opposite of -3 is 3.

$$x=\frac{3±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±1}{-2}$$

Now solve the equation x=\frac{3±1}{-2} when ± is plus. Add 3 to 1.

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

Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{3±1}{-2} when ± is minus. Subtract 1 from 3.

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

Divide 2 by -2.

$$x=-1$$

The equation is now solved.

$$x=-2 x=-1$$

Steps for Completing the Square

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

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

To find the opposite of 4+4x+x^{2}, find the opposite of each term.

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

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

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

Add 4 to both sides.

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

Add -2 and 4 to get 2.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -3 by -1.

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

Divide 2 by -1.

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

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.

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

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

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

Add -2 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{1}{4}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=-1 x=-2$$