Solve Using the Quadratic Formula 16t^2-144t+288=0

$16 t^{2} - 144 t + 288 = 0$

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 $a = 16$ , $b = - 144$ , and $c = 288$ into the quadratic formula and solve for $t$ .

$\frac{144 \pm \sqrt{{(- 144)}^{2} - 4 \cdot (16 \cdot 288)}}{2 \cdot 16}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 144$ to the power of $2$ .

$t = \frac{144 \pm \sqrt{20736 - 4 \cdot 16 \cdot 288}}{2 \cdot 16}$

Multiply $- 4 \cdot 16 \cdot 288$ .

Multiply $- 4$ by $16$ .

$t = \frac{144 \pm \sqrt{20736 - 64 \cdot 288}}{2 \cdot 16}$

Multiply $- 64$ by $288$ .

$t = \frac{144 \pm \sqrt{20736 - 18432}}{2 \cdot 16}$

Subtract $18432$ from $20736$ .

$t = \frac{144 \pm \sqrt{2304}}{2 \cdot 16}$

Rewrite $2304$ as $48^{2}$ .

$t = \frac{144 \pm \sqrt{48^{2}}}{2 \cdot 16}$

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

$t = \frac{144 \pm 48}{2 \cdot 16}$

Multiply $2$ by $16$ .

$t = \frac{144 \pm 48}{32}$

Simplify $\frac{144 \pm 48}{32}$ .

$t = \frac{9 \pm 3}{2}$

Step 4

The final answer is the combination of both solutions.

$t = 6, 3$