Solve 4-x^2=2+x

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

Solve for x

$$x=-2$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 2 from both sides.

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

Subtract 2 from 4 to get 2.

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

Subtract x from both sides.

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

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

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

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.

$$a+b=-1 ab=-2=-2$$

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.

$$a=1 b=-2$$

Rewrite -x^{2}-x+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

Subtract 2 from both sides.

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

Subtract 2 from 4 to get 2.

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

Subtract x from both sides.

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

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.

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

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

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

Multiply -4 times -1.

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

Multiply 4 times 2.

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

Add 1 to 8.

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

Take the square root of 9.

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

The opposite of -1 is 1.

$$x=\frac{1±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±3}{-2}$$

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

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

Divide 4 by -2.

$$x=-2$$

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

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-2 x=1$$

Steps for Completing the Square

Subtract x from both sides.

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

Subtract 4 from both sides.

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

Subtract 4 from 2 to get -2.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -1 by -1.

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

Divide -2 by -1.

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

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.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=1 x=-2$$