Precalculus Examples

$14 = 2 + 29 t - 16 t^{2}$

Step 1

Rewrite the equation as $2 + 29 t - 16 t^{2} = 14$ .

$2 + 29 t - 16 t^{2} = 14$

Step 2

Move all terms to the left side of the equation and simplify.

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Step 2.1

Subtract $14$ from both sides of the equation.

$2 + 29 t - 16 t^{2} - 14 = 0$

Step 2.2

Subtract $14$ from $2$ .

$29 t - 16 t^{2} - 12 = 0$

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 $c = - 12$ into the quadratic formula and solve for $t$ .

$\frac{- 29 \pm \sqrt{29^{2} - 4 \cdot (- 16 \cdot - 12)}}{2 \cdot - 16}$

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 $2$ .

$t = \frac{- 29 \pm \sqrt{841 - 4 \cdot - 16 \cdot - 12}}{2 \cdot - 16}$

Step 5.1.2

Multiply $- 4 \cdot - 16 \cdot - 12$ .

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Step 5.1.2.1

Multiply $- 4$ by $- 16$ .

$t = \frac{- 29 \pm \sqrt{841 + 64 \cdot - 12}}{2 \cdot - 16}$

Step 5.1.2.2

Multiply $64$ by $- 12$ .

$t = \frac{- 29 \pm \sqrt{841 - 768}}{2 \cdot - 16}$

Step 5.1.3

Subtract $768$ from $841$ .

$t = \frac{- 29 \pm \sqrt{73}}{2 \cdot - 16}$

Step 5.2

Multiply $2$ by $- 16$ .

$t = \frac{- 29 \pm \sqrt{73}}{- 32}$

Step 5.3

Simplify $\frac{- 29 \pm \sqrt{73}}{- 32}$ .

$t = \frac{29 \pm \sqrt{73}}{32}$

Step 6

The final answer is the combination of both solutions.

$t = \frac{29 + \sqrt{73}}{32}, \frac{29 - \sqrt{73}}{32}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$t = \frac{29 + \sqrt{73}}{32}, \frac{29 - \sqrt{73}}{32}$

Decimal Form:

$t = 1.17325011 \dots, 0.63924988 \dots$