Calculus Examples

$10 p^{2} d^{2} + 7 p d t = 12 t^{2}$

Step 1

Subtract $12 t^{2}$ from both sides of the equation.

$10 p^{2} d^{2} + 7 p d t - 12 t^{2} = 0$

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 $a = - 12$ , $b = 7 p d$ , and $c = 10 p^{2} d^{2}$ into the quadratic formula and solve for $t$ .

$\frac{- 7 p d \pm \sqrt{{(7 p d)}^{2} - 4 \cdot (- 12 \cdot (10 p^{2} d^{2}))}}{2 \cdot - 12}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 4.1.1.1

Apply the product rule to $7 p d$ .

$t = \frac{- 7 p d \pm \sqrt{{(7 p)}^{2} d^{2} - 4 \cdot - 12 \cdot (10 p^{2} d^{2})}}{2 \cdot - 12}$

Step 4.1.1.2

Apply the product rule to $7 p$ .

$t = \frac{- 7 p d \pm \sqrt{7^{2} p^{2} d^{2} - 4 \cdot - 12 \cdot (10 p^{2} d^{2})}}{2 \cdot - 12}$

Step 4.1.2

Raise $7$ to the power of $2$ .

$t = \frac{- 7 p d \pm \sqrt{49 p^{2} d^{2} - 4 \cdot - 12 \cdot (10 p^{2} d^{2})}}{2 \cdot - 12}$

Step 4.1.3

Multiply $- 4 \cdot - 12 \cdot 10$ .

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Step 4.1.3.1

Multiply $- 4$ by $- 12$ .

$t = \frac{- 7 p d \pm \sqrt{49 p^{2} d^{2} + 48 \cdot (10 p^{2} d^{2})}}{2 \cdot - 12}$

Step 4.1.3.2

Multiply $48$ by $10$ .

$t = \frac{- 7 p d \pm \sqrt{49 p^{2} d^{2} + 480 p^{2} d^{2}}}{2 \cdot - 12}$

Step 4.1.4

Add $49 p^{2} d^{2}$ and $480 p^{2} d^{2}$ .

$t = \frac{- 7 p d \pm \sqrt{529 p^{2} d^{2}}}{2 \cdot - 12}$

Step 4.1.5

Rewrite $529 p^{2} d^{2}$ as ${(23 p d)}^{2}$ .

$t = \frac{- 7 p d \pm \sqrt{{(23 p d)}^{2}}}{2 \cdot - 12}$

Step 4.1.6

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

$t = \frac{- 7 p d \pm 23 p d}{2 \cdot - 12}$

Step 4.2

Multiply $2$ by $- 12$ .

$t = \frac{- 7 p d \pm 23 p d}{- 24}$

Step 4.3

Simplify $\frac{- 7 p d \pm 23 p d}{- 24}$ .

$t = \frac{7 p d \pm 23 p d}{24}$

Step 5

The final answer is the combination of both solutions.

$t = \frac{5 p d}{4}$

$t = - \frac{2 p d}{3}$