Solve for t 240=-16t^2+128t

$240 = - 16 t^{2} + 128 t$

Step 1

Rewrite the equation as $- 16 t^{2} + 128 t = 240$ .

$- 16 t^{2} + 128 t = 240$

Step 2

Subtract $240$ from both sides of the equation.

$- 16 t^{2} + 128 t - 240 = 0$

Step 3

Factor the left side of the equation.

Factor $- 16$ out of $- 16 t^{2} + 128 t - 240$ .

Factor $- 16$ out of $- 16 t^{2}$ .

$- 16 t^{2} + 128 t - 240 = 0$

Factor $- 16$ out of $128 t$ .

$- 16 t^{2} - 16 (- 8 t) - 240 = 0$

Factor $- 16$ out of $- 240$ .

$- 16 t^{2} - 16 (- 8 t) - 16 \cdot 15 = 0$

Factor $- 16$ out of $- 16 (t^{2}) - 16 (- 8 t)$ .

$- 16 (t^{2} - 8 t) - 16 \cdot 15 = 0$

Factor $- 16$ out of $- 16 (t^{2} - 8 t) - 16 (15)$ .

$- 16 (t^{2} - 8 t + 15) = 0$

Factor.

Factor $t^{2} - 8 t + 15$ 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 $15$ and whose sum is $- 8$ .

$- 5, - 3$

Write the factored form using these integers.

$- 16 ((t - 5) (t - 3)) = 0$

Remove unnecessary parentheses.

$- 16 (t - 5) (t - 3) = 0$

Step 4

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

$t - 5 = 0$

$t - 3 = 0$

Step 5

Set $t - 5$ equal to $0$ and solve for $t$ .

Set $t - 5$ equal to $0$ .

$t - 5 = 0$

Add $5$ to both sides of the equation.

$t = 5$

Step 6

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

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

$t - 3 = 0$

Add $3$ to both sides of the equation.

$t = 3$

Step 7

The final solution is all the values that make $- 16 (t - 5) (t - 3) = 0$ true.

$t = 5, 3$

Solve for t 240=-16t^2+128t

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