Solve for n 2n^2+4n-16=0

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

Step 1

Factor the left side of the equation.

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

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

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

Factor $2$ out of $4 n$ .

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

Factor $2$ out of $- 16$ .

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

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

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

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

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

Factor.

Factor $n^{2} + 2 n - 8$ 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.

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

Remove unnecessary parentheses.

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

Step 2

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

$n - 2 = 0$

$n + 4 = 0$

Step 3

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

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

$n - 2 = 0$

Add $2$ to both sides of the equation.

$n = 2$

Step 4

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

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

$n + 4 = 0$

Subtract $4$ from both sides of the equation.

$n = - 4$

Step 5

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

$n = 2, - 4$

Solve for n 2n^2+4n-16=0

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