Solve x^2=2x+8

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

Solve for x

$$x=-2$$

$$x=4$$

Steps Using Factoring

Subtract 2x from both sides.

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

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-4 b=2$$

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

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

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

$$x=4 x=-2$$

Steps Using Factoring By Grouping

Subtract 2x from both sides.

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

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-4 b=2$$

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

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

Factor out x in the first and 2 in the second group.

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

Factor out common term x-4 by using distributive property.

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

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

$$x=4 x=-2$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

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}.

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

Square -2.

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

Multiply -4 times -8.

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

Add 4 to 32.

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

Take the square root of 36.

$$x=\frac{-\left(-2\right)±6}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±6}{2}$$

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

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

Divide 8 by 2.

$$x=4$$

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

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=4 x=-2$$

Steps for Completing the Square

Subtract 2x from both sides.

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

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.

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

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.

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

Add 1 to both sides of the equation.

$$x=4 x=-2$$