Solve Using the Quadratic Formula 16t^2-96t+128=0

$16 t^{2} - 96 t + 128 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 16$ , $b = - 96$ , and $c = 128$ into the quadratic formula and solve for $t$ .

$\frac{96 \pm \sqrt{{(- 96)}^{2} - 4 \cdot (16 \cdot 128)}}{2 \cdot 16}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 96$ to the power of $2$ .

$t = \frac{96 \pm \sqrt{9216 - 4 \cdot 16 \cdot 128}}{2 \cdot 16}$

Multiply $- 4 \cdot 16 \cdot 128$ .

Multiply $- 4$ by $16$ .

$t = \frac{96 \pm \sqrt{9216 - 64 \cdot 128}}{2 \cdot 16}$

Multiply $- 64$ by $128$ .

$t = \frac{96 \pm \sqrt{9216 - 8192}}{2 \cdot 16}$

Subtract $8192$ from $9216$ .

$t = \frac{96 \pm \sqrt{1024}}{2 \cdot 16}$

Rewrite $1024$ as $32^{2}$ .

$t = \frac{96 \pm \sqrt{32^{2}}}{2 \cdot 16}$

Pull terms out from under the radical, assuming positive real numbers.

$t = \frac{96 \pm 32}{2 \cdot 16}$

Multiply $2$ by $16$ .

$t = \frac{96 \pm 32}{32}$

Simplify $\frac{96 \pm 32}{32}$ .

$t = 3 \pm 1$

Step 4

The final answer is the combination of both solutions.

$t = 4, 2$