Solve 0=-2left(x+1right)^2+8

$$0=-2 { \left(x+1 \right) }^{ 2 } +8$$

Solve for x

$$x=1$$

$$x=-3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$0=-2\left(x^{2}+2x+1\right)+8$$

Use the distributive property to multiply -2 by x^{2}+2x+1.

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

Add -2 and 8 to get 6.

$$0=-2x^{2}-4x+6$$

Swap sides so that all variable terms are on the left hand side.

$$-2x^{2}-4x+6=0$$

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3=-3$$

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.

$$a=1 b=-3$$

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

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

Factor out x in the first and 3 in the second group.

$$x\left(-x+1\right)+3\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(x+3\right)$$

To find equation solutions, solve -x+1=0 and x+3=0.

$$x=1 x=-3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$0=-2\left(x^{2}+2x+1\right)+8$$

Use the distributive property to multiply -2 by x^{2}+2x+1.

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

Add -2 and 8 to get 6.

$$0=-2x^{2}-4x+6$$

Swap sides so that all variable terms are on the left hand side.

$$-2x^{2}-4x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -4 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+8\times 6}}{2\left(-2\right)}$$

Multiply 8 times 6.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\left(-2\right)}$$

Add 16 to 48.

$$x=\frac{-\left(-4\right)±\sqrt{64}}{2\left(-2\right)}$$

Take the square root of 64.

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

The opposite of -4 is 4.

$$x=\frac{4±8}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{4±8}{-4}$$

Now solve the equation x=\frac{4±8}{-4} when ± is plus. Add 4 to 8.

$$x=\frac{12}{-4}$$

Divide 12 by -4.

$$x=-3$$

Now solve the equation x=\frac{4±8}{-4} when ± is minus. Subtract 8 from 4.

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

Divide -4 by -4.

$$x=1$$

The equation is now solved.

$$x=-3 x=1$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$0=-2\left(x^{2}+2x+1\right)+8$$

Use the distributive property to multiply -2 by x^{2}+2x+1.

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

Add -2 and 8 to get 6.

$$0=-2x^{2}-4x+6$$

Swap sides so that all variable terms are on the left hand side.

$$-2x^{2}-4x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}-4x=-6$$

Divide both sides by -2.

$$\frac{-2x^{2}-4x}{-2}=-\frac{6}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{4}{-2}\right)x=-\frac{6}{-2}$$

Divide -4 by -2.

$$x^{2}+2x=-\frac{6}{-2}$$

Divide -6 by -2.

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

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

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square 1.

$$x^{2}+2x+1=3+1$$

Add 3 to 1.

$$x^{2}+2x+1=4$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$