Find the Roots (Zeros) 2x^2+4x=16

$2 x^{2} + 4 x = 16$

Step 1

Subtract $16$ from both sides of the equation.

$2 x^{2} + 4 x - 16 = 0$

Step 2

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 4 x - 16$ .

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) + 4 x - 16 = 0$

Factor $2$ out of $4 x$ .

$2 (x^{2}) + 2 (2 x) - 16 = 0$

Factor $2$ out of $- 16$ .

$2 x^{2} + 2 (2 x) + 2 \cdot - 8 = 0$

Factor $2$ out of $2 x^{2} + 2 (2 x)$ .

$2 (x^{2} + 2 x) + 2 \cdot - 8 = 0$

Factor $2$ out of $2 (x^{2} + 2 x) + 2 \cdot - 8$ .

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

Factor.

Factor $x^{2} + 2 x - 8$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $2$ .

$- 2, 4$

Write the factored form using these integers.

$2 ((x - 2) (x + 4)) = 0$

Remove unnecessary parentheses.

$2 (x - 2) (x + 4) = 0$

Step 3

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

$x - 2 = 0$

$x + 4 = 0$

Step 4

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 5

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 6

The final solution is all the values that make $2 (x - 2) (x + 4) = 0$ true.

$x = 2, - 4$

Step 7

Find the Roots (Zeros) 2x^2+4x=16

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