Solve x^2-2x=4-x^2

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

Solve for x

$$x=-1$$

$$x=2$$

Steps Using Factoring By Grouping

Subtract 4 from both sides.

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

Add x^{2} to both sides.

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Divide both sides by 2.

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

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.

$$a+b=-1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

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

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

Factor out x in x^{2}-2x.

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

Factor out common term x-2 by using distributive property.

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

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

$$x=2 x=-1$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

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

Add x^{2} to both sides.

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

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

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

Square -2.

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

Multiply -4 times 2.

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

Multiply -8 times -4.

$$x=\frac{-\left(-2\right)±\sqrt{4+32}}{2\times 2}$$

Add 4 to 32.

$$x=\frac{-\left(-2\right)±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

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

The opposite of -2 is 2.

$$x=\frac{2±6}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2±6}{4}$$

Now solve the equation x=\frac{2±6}{4} when ± is plus. Add 2 to 6.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{2±6}{4} when ± is minus. Subtract 6 from 2.

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

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

$$x=2 x=-1$$

Steps for Completing the Square

Add x^{2} to both sides.

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

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

Divide 4 by 2.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{1}{4}.

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

Factor x^{2}-x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$x=2 x=-1$$