Finite Math Examples

$h = 4 + 29 t - 16 t^{2}$

Step 1

Rewrite the equation as $4 + 29 t - 16 t^{2} = h$ .

$4 + 29 t - 16 t^{2} = h$

Step 2

Subtract $h$ from both sides of the equation.

$4 + 29 t - 16 t^{2} - h = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = - 16$ , $b = 29$ , and $c = 4 - h$ into the quadratic formula and solve for $t$ .

$\frac{- 29 \pm \sqrt{29^{2} - 4 \cdot (- 16 \cdot (4 - h))}}{2 \cdot - 16}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $29$ to the power of $2$ .

$t = \frac{- 29 \pm \sqrt{841 - 4 \cdot - 16 \cdot (4 - h)}}{2 \cdot - 16}$

Step 5.1.2

Multiply $- 4$ by $- 16$ .

$t = \frac{- 29 \pm \sqrt{841 + 64 \cdot (4 - h)}}{2 \cdot - 16}$

Step 5.1.3

Apply the distributive property.

$t = \frac{- 29 \pm \sqrt{841 + 64 \cdot 4 + 64 (- h)}}{2 \cdot - 16}$

Step 5.1.4

Multiply $64$ by $4$ .

$t = \frac{- 29 \pm \sqrt{841 + 256 + 64 (- h)}}{2 \cdot - 16}$

Step 5.1.5

Multiply $- 1$ by $64$ .

$t = \frac{- 29 \pm \sqrt{841 + 256 - 64 h}}{2 \cdot - 16}$

Step 5.1.6

Add $841$ and $256$ .

$t = \frac{- 29 \pm \sqrt{1097 - 64 h}}{2 \cdot - 16}$

Step 5.2

Multiply $2$ by $- 16$ .

$t = \frac{- 29 \pm \sqrt{1097 - 64 h}}{- 32}$

Step 5.3

Simplify $\frac{- 29 \pm \sqrt{1097 - 64 h}}{- 32}$ .

$t = \frac{29 \pm \sqrt{1097 - 64 h}}{32}$

Step 6

The final answer is the combination of both solutions.

$t = \frac{29 + \sqrt{1097 - 64 h}}{32}$

$t = \frac{29 - \sqrt{1097 - 64 h}}{32}$