Solve by Factoring x^2+3x=18

$x^{2} + 3 x = 18$

Step 1

Subtract $18$ from both sides of the equation.

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

Step 2

Factor $x^{2} + 3 x - 18$ 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.

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

Step 3

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

$x - 3 = 0$

$x + 6 = 0$

Step 4

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 5

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

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

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 6

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

$x = 3, - 6$

Solve by Factoring x^2+3x=18

Related Questions

Support LCMGCF.com by disabling your adblocker.