Solve Using the Square Root Property n^2+6n=40

$n^{2} + 6 n = 40$

Step 1

Subtract $40$ from both sides of the equation.

$n^{2} + 6 n - 40 = 0$

Step 2

Factor $n^{2} + 6 n - 40$ 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 $- 40$ and whose sum is $6$ .

$- 4, 10$

Write the factored form using these integers.

$(n - 4) (n + 10) = 0$

Step 3

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

$n - 4 = 0$

$n + 10 = 0$

Step 4

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

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

$n - 4 = 0$

Add $4$ to both sides of the equation.

$n = 4$

Step 5

Set $n + 10$ equal to $0$ and solve for $n$ .

Set $n + 10$ equal to $0$ .

$n + 10 = 0$

Subtract $10$ from both sides of the equation.

$n = - 10$

Step 6

The final solution is all the values that make $(n - 4) (n + 10) = 0$ true.

$n = 4, - 10$

Solve Using the Square Root Property n^2+6n=40

Related Questions