Solve -2x^2+4/4-x^2+2x+8/4-x^2=0

$$- \frac { 2 x ^ { 2 } + 4 } { 4 - x ^ { 2 } } + \frac { 2 x + 8 } { 4 - x ^ { 2 } } = 0$$

Solve for x

$$x=-1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(-x-2\right).

$$-\left(2x^{2}+4\right)+2x+8=0$$

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

$$-2x^{2}-4+2x+8=0$$

Add -4 and 8 to get 4.

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

Divide both sides by 2.

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

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 positive, the positive number has greater absolute value than the negative. 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 the first and -1 in the second group.

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

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.

$$x=2 x=-1$$

Variable x cannot be equal to 2.

$$x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(-x-2\right).

$$-\left(2x^{2}+4\right)+2x+8=0$$

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

$$-2x^{2}-4+2x+8=0$$

Add -4 and 8 to get 4.

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

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.

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

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}.

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

Square 2.

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

Multiply -4 times -2.

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

Multiply 8 times 4.

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

Add 4 to 32.

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

Take the square root of 36.

$$x=\frac{-2±6}{2\left(-2\right)}$$

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.

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

Divide 4 by -4.

$$x=-1$$

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

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

Divide -8 by -4.

$$x=2$$

The equation is now solved.

$$x=-1 x=2$$

Variable x cannot be equal to 2.

$$x=-1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(-x-2\right).

$$-\left(2x^{2}+4\right)+2x+8=0$$

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

$$-2x^{2}-4+2x+8=0$$

Add -4 and 8 to get 4.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

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.

$$x=2 x=-1$$

Variable x cannot be equal to 2.

$$x=-1$$