Solve 8=x^2+2x

$$8 = x ^ { 2 } + 2 x$$

Solve for x

$$x=-4$$

$$x=2$$

Steps Using Factoring

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

$$x^{2}+2x=8$$

Subtract 8 from both sides.

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

To solve the equation, factor x^{2}+2x-8 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.

$$a+b=2 ab=-8$$

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. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

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

$$\left(x-2\right)\left(x+4\right)$$

To find equation solutions, solve x-2=0 and x+4=0.

$$x=2 x=-4$$

Steps Using Factoring By Grouping

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

$$x^{2}+2x=8$$

Subtract 8 from both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-8\right)=-8$$

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. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

Rewrite x^{2}+2x-8 as \left(x^{2}-2x\right)+\left(4x-8\right).

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

Factor out x in the first and 4 in the second group.

$$x\left(x-2\right)+4\left(x-2\right)$$

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

$$\left(x-2\right)\left(x+4\right)$$

To find equation solutions, solve x-2=0 and x+4=0.

$$x=2 x=-4$$

Steps Using the Quadratic Formula

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

$$x^{2}+2x=8$$

Subtract 8 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 2.

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

Multiply -4 times -8.

$$x=\frac{-2±\sqrt{4+32}}{2}$$

Add 4 to 32.

$$x=\frac{-2±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-2±6}{2}$$

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

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

Divide 4 by 2.

$$x=2$$

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

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

The equation is now solved.

$$x=2 x=-4$$

Steps for Completing the Square

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

$$x^{2}+2x=8$$

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.

$$x^{2}+2x+1^{2}=8+1^{2}$$

Square 1.

$$x^{2}+2x+1=8+1$$

Add 8 to 1.

$$x^{2}+2x+1=9$$

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}.

$$\left(x+1\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+1=3 x+1=-3$$

Subtract 1 from both sides of the equation.

$$x=2 x=-4$$