Solve by Factoring x^2+2x-4=2x

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

Step 1

Subtract $2 x$ from both sides of the equation.

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

Step 2

Combine the opposite terms in $x^{2} + 2 x - 4 - 2 x$ .

Subtract $2 x$ from $2 x$ .

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

Add $x^{2}$ and $0$ .

$x^{2} - 4 = 0$

Step 3

Rewrite $4$ as $2^{2}$ .

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

Step 4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$(x + 2) (x - 2) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 2 = 0$

$x - 2 = 0$

Step 6

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 8

The final solution is all the values that make $(x + 2) (x - 2) = 0$ true.

$x = - 2, 2$