Solve by Factoring n^2+3n-12=6

$n^{2} + 3 n - 12 = 6$

Step 1

Subtract $6$ from both sides of the equation.

$n^{2} + 3 n - 12 - 6 = 0$

Step 2

Subtract $6$ from $- 12$ .

$n^{2} + 3 n - 18 = 0$

Step 3

Factor $n^{2} + 3 n - 18$ 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 $- 18$ and whose sum is $3$ .

$- 3, 6$

Write the factored form using these integers.

$(n - 3) (n + 6) = 0$

Step 4

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

$n - 3 = 0$

$n + 6 = 0$

Step 5

Set $n - 3$ equal to $0$ and solve for $n$ .

Set $n - 3$ equal to $0$ .

$n - 3 = 0$

Add $3$ to both sides of the equation.

$n = 3$

Step 6

Set $n + 6$ equal to $0$ and solve for $n$ .

Set $n + 6$ equal to $0$ .

$n + 6 = 0$

Subtract $6$ from both sides of the equation.

$n = - 6$

Step 7

The final solution is all the values that make $(n - 3) (n + 6) = 0$ true.

$n = 3, - 6$

Solve by Factoring n^2+3n-12=6

Related Questions

Support LCMGCF.com by disabling your adblocker.