Solve 2left(x+3right)^2=8

$$2 { \left(x+3 \right) }^{ 2 } = 8$$

Solve for x

$$x=-1$$

$$x=-5$$

Steps Using Factoring

Divide both sides by 2.

$$\left(x+3\right)^{2}=\frac{8}{2}$$

Divide 8 by 2 to get 4.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=4$$

Subtract 4 from both sides.

$$x^{2}+6x+9-4=0$$

Subtract 4 from 9 to get 5.

$$x^{2}+6x+5=0$$

To solve the equation, factor x^{2}+6x+5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=6 ab=5$$

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

$$a=1 b=5$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x+1\right)\left(x+5\right)$$

To find equation solutions, solve x+1=0 and x+5=0.

$$x=-1 x=-5$$

Steps Using Factoring By Grouping

Divide both sides by 2.

$$\left(x+3\right)^{2}=\frac{8}{2}$$

Divide 8 by 2 to get 4.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=4$$

Subtract 4 from both sides.

$$x^{2}+6x+9-4=0$$

Subtract 4 from 9 to get 5.

$$x^{2}+6x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\times 5=5$$

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

$$a=1 b=5$$

Rewrite x^{2}+6x+5 as \left(x^{2}+x\right)+\left(5x+5\right).

$$\left(x^{2}+x\right)+\left(5x+5\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x+1\right)+5\left(x+1\right)$$

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

$$\left(x+1\right)\left(x+5\right)$$

To find equation solutions, solve x+1=0 and x+5=0.

$$x=-1 x=-5$$

Steps Using the Quadratic Formula

Divide both sides by 2.

$$\left(x+3\right)^{2}=\frac{8}{2}$$

Divide 8 by 2 to get 4.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=4$$

Subtract 4 from both sides.

$$x^{2}+6x+9-4=0$$

Subtract 4 from 9 to get 5.

$$x^{2}+6x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-6±\sqrt{36-20}}{2}$$

Add 36 to -20.

$$x=\frac{-6±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-6±4}{2}$$

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

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

Divide -2 by 2.

$$x=-1$$

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

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=-1 x=-5$$

Steps for Completing the Square

Divide both sides by 2.

$$\left(x+3\right)^{2}=\frac{8}{2}$$

Divide 8 by 2 to get 4.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+3=2 x+3=-2$$

Subtract 3 from both sides of the equation.

$$x=-1 x=-5$$