Solve for t t^2+15t=-36

$t^{2} + 15 t = - 36$

Step 1

Add $36$ to both sides of the equation.

$t^{2} + 15 t + 36 = 0$

Step 2

Factor $t^{2} + 15 t + 36$ 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 $36$ and whose sum is $15$ .

$3, 12$

Write the factored form using these integers.

$(t + 3) (t + 12) = 0$

Step 3

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

$t + 3 = 0$

$t + 12 = 0$

Step 4

Set $t + 3$ equal to $0$ and solve for $t$ .

Set $t + 3$ equal to $0$ .

$t + 3 = 0$

Subtract $3$ from both sides of the equation.

$t = - 3$

Step 5

Set $t + 12$ equal to $0$ and solve for $t$ .

Set $t + 12$ equal to $0$ .

$t + 12 = 0$

Subtract $12$ from both sides of the equation.

$t = - 12$

Step 6

The final solution is all the values that make $(t + 3) (t + 12) = 0$ true.

$t = - 3, - 12$

Solve for t t^2+15t=-36

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