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

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

Step 1

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

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

Step 2

Subtract $128$ from both sides of the equation.

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

Step 3

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

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

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

Factor $16$ out of $128 t$ .

$16 (t^{2}) + 16 (8 t) - 128 = 0$

Factor $16$ out of $- 128$ .

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

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

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

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

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

Step 4

Divide each term in $16 (t^{2} + 8 t - 8) = 0$ by $16$ and simplify.

Divide each term in $16 (t^{2} + 8 t - 8) = 0$ by $16$ .

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

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

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

Divide $t^{2} + 8 t - 8$ by $1$ .

$t^{2} + 8 t - 8 = \frac{0}{16}$

Simplify the right side.

Divide $0$ by $16$ .

$t^{2} + 8 t - 8 = 0$

Step 5

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 6

Substitute the values $a = 1$ , $b = 8$ , and $c = - 8$ into the quadratic formula and solve for $t$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (1 \cdot - 8)}}{2 \cdot 1}$

Step 7

Simplify.

Simplify the numerator.

Raise $8$ to the power of $2$ .

$t = \frac{- 8 \pm \sqrt{64 - 4 \cdot 1 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 8$ .

Multiply $- 4$ by $1$ .

$t = \frac{- 8 \pm \sqrt{64 - 4 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4$ by $- 8$ .

$t = \frac{- 8 \pm \sqrt{64 + 32}}{2 \cdot 1}$

Add $64$ and $32$ .

$t = \frac{- 8 \pm \sqrt{96}}{2 \cdot 1}$

Rewrite $96$ as $4^{2} \cdot 6$ .

Factor $16$ out of $96$ .

$t = \frac{- 8 \pm \sqrt{16 (6)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$t = \frac{- 8 \pm \sqrt{4^{2} \cdot 6}}{2 \cdot 1}$

Pull terms out from under the radical.

$t = \frac{- 8 \pm 4 \sqrt{6}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$t = \frac{- 8 \pm 4 \sqrt{6}}{2}$

Simplify $\frac{- 8 \pm 4 \sqrt{6}}{2}$ .

$t = - 4 \pm 2 \sqrt{6}$

Step 8

The final answer is the combination of both solutions.

$t = - 4 + 2 \sqrt{6}, - 4 - 2 \sqrt{6}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$t = - 4 + 2 \sqrt{6}, - 4 - 2 \sqrt{6}$

Decimal Form:

$t = 0.89897948 \dots, - 8.89897948 \dots$

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

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