Basic Math Examples

$d = v t + \frac{1}{2} \cdot (a t^{2})$

Step 1

Rewrite the equation as

$v t + \frac{1}{2} \cdot (a t^{2}) = d$ .

$v t + \frac{1}{2} \cdot (a t^{2}) = d$

Step 2

Multiply

$\frac{1}{2} (a t^{2})$ .

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Step 2.1

Combine

$a$ and

$\frac{1}{2}$ .

$v t + \frac{a}{2} t^{2} = d$

Step 2.2

Combine

$\frac{a}{2}$ and

$t^{2}$ .

$v t + \frac{a t^{2}}{2} = d$

Step 3

Subtract

$d$ from both sides of the equation.

$v t + \frac{a t^{2}}{2} - d = 0$

Step 4

Multiply through by the least common denominator

$2$ , then simplify.

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Step 4.1

Apply the distributive property.

$2 (v t) + 2 (\frac{a t^{2}}{2}) + 2 (- d) = 0$

Step 4.2

Simplify.

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Step 4.2.1

Cancel the common factor of

$2$ .

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Step 4.2.1.1

Cancel the common factor.

$2 v t + 2 (\frac{a t^{2}}{2}) + 2 (- d) = 0$

Step 4.2.1.2

Rewrite the expression.

$2 v t + a t^{2} + 2 (- d) = 0$

Step 4.2.2

Multiply

$- 1$ by

$2$ .

$2 v t + a t^{2} - 2 d = 0$

Step 4.3

Move

$v$ .

$2 t v + a t^{2} - 2 d = 0$

Step 4.4

Reorder

$2 t v$ and

$a t^{2}$ .

$a t^{2} + 2 t v - 2 d = 0$

Step 5

Use the quadratic formula to find the solutions.

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

Step 6

Substitute the values

$a = a$ ,

$b = 2 v$ , and

$c = - 2 d$ into the quadratic formula and solve for

$t$ .

$\frac{- 2 v \pm \sqrt{{(2 v)}^{2} - 4 \cdot (a \cdot (- 2 d))}}{2 a}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Add parentheses.

$t = \frac{- 2 v \pm \sqrt{{(2 v)}^{2} - 4 \cdot (a \cdot (- 2 d))}}{2 a}$

Step 7.1.2

Let

$u = a \cdot (- 2 d)$ . Substitute

$u$ for all occurrences of

$a \cdot (- 2 d)$ .

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Step 7.1.2.1

Apply the product rule to

$2 v$ .

$t = \frac{- 2 v \pm \sqrt{2^{2} v^{2} - 4 \cdot u}}{2 a}$

Step 7.1.2.2

Raise

$2$ to the power of

$2$ .

$t = \frac{- 2 v \pm \sqrt{4 v^{2} - 4 u}}{2 a}$

Step 7.1.3

Factor

$4$ out of

$4 v^{2} - 4 u$ .

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Step 7.1.3.1

Factor

$4$ out of

$4 v^{2}$ .

$t = \frac{- 2 v \pm \sqrt{4 (v^{2}) - 4 u}}{2 a}$

Step 7.1.3.2

Factor

$4$ out of

$- 4 u$ .

$t = \frac{- 2 v \pm \sqrt{4 (v^{2}) + 4 (- u)}}{2 a}$

Step 7.1.3.3

Factor

$4$ out of

$4 (v^{2}) + 4 (- u)$ .

$t = \frac{- 2 v \pm \sqrt{4 (v^{2} - u)}}{2 a}$

Step 7.1.4

Replace all occurrences of

$u$ with

$a \cdot (- 2 d)$ .

$t = \frac{- 2 v \pm \sqrt{4 (v^{2} - (a \cdot (- 2 d)))}}{2 a}$

Step 7.1.5

Simplify each term.

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Step 7.1.5.1

Rewrite using the commutative property of multiplication.

$t = \frac{- 2 v \pm \sqrt{4 (v^{2} - (- 2 a d))}}{2 a}$

Step 7.1.5.2

Multiply

$- 2$ by

$- 1$ .

$t = \frac{- 2 v \pm \sqrt{4 (v^{2} + 2 a d)}}{2 a}$

Step 7.1.6

Rewrite

$4$ as

$2^{2}$ .

$t = \frac{- 2 v \pm \sqrt{2^{2} (v^{2} + 2 a d)}}{2 a}$

Step 7.1.7

Pull terms out from under the radical.

$t = \frac{- 2 v \pm 2 \sqrt{v^{2} + 2 a d}}{2 a}$

Step 7.2

Simplify

$\frac{- 2 v \pm 2 \sqrt{v^{2} + 2 a d}}{2 a}$ .

$t = \frac{- v \pm \sqrt{v^{2} + 2 a d}}{a}$

Step 8

The final answer is the combination of both solutions.

$t = - \frac{v - \sqrt{v^{2} + 2 a d}}{a}$

$t = - \frac{v + \sqrt{v^{2} + 2 a d}}{a}$