Algebra Examples

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

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Add $16 t^{2}$ to both sides of the equation.

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

Step 1.2

Subtract $128 t$ from both sides of the equation.

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

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 16$ , $b = - 128$ , and $c = 240$ into the quadratic formula and solve for $t$ .

$\frac{128 \pm \sqrt{{(- 128)}^{2} - 4 \cdot (16 \cdot 240)}}{2 \cdot 16}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 128$ to the power of $2$ .

$t = \frac{128 \pm \sqrt{16384 - 4 \cdot 16 \cdot 240}}{2 \cdot 16}$

Step 4.1.2

Multiply $- 4 \cdot 16 \cdot 240$ .

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Step 4.1.2.1

Multiply $- 4$ by $16$ .

$t = \frac{128 \pm \sqrt{16384 - 64 \cdot 240}}{2 \cdot 16}$

Step 4.1.2.2

Multiply $- 64$ by $240$ .

$t = \frac{128 \pm \sqrt{16384 - 15360}}{2 \cdot 16}$

Step 4.1.3

Subtract $15360$ from $16384$ .

$t = \frac{128 \pm \sqrt{1024}}{2 \cdot 16}$

Step 4.1.4

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

$t = \frac{128 \pm \sqrt{32^{2}}}{2 \cdot 16}$

Step 4.1.5

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

$t = \frac{128 \pm 32}{2 \cdot 16}$

Step 4.2

Multiply $2$ by $16$ .

$t = \frac{128 \pm 32}{32}$

Step 4.3

Simplify $\frac{128 \pm 32}{32}$ .

$t = 4 \pm 1$

Step 5

The final answer is the combination of both solutions.

$t = 5, 3$