Precalculus Examples

$a \cdot 12 = n (n - 6)$

Step 1

Rewrite the equation as $n (n - 6) = a \cdot 12$ .

$n (n - 6) = a \cdot 12$

Step 2

Simplify $n (n - 6)$ .

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Step 2.1

Apply the distributive property.

$n \cdot n + n \cdot - 6 = a \cdot 12$

Step 2.2

Simplify the expression.

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Step 2.2.1

Multiply $n$ by $n$ .

$n^{2} + n \cdot - 6 = a \cdot 12$

Step 2.2.2

Move $- 6$ to the left of $n$ .

$n^{2} - 6 n = a \cdot 12$

Step 3

Move $12$ to the left of $a$ .

$n^{2} - 6 n = 12 a$

Step 4

Subtract $12 a$ from both sides of the equation.

$n^{2} - 6 n - 12 a = 0$

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 $a = 1$ , $b = - 6$ , and $c = - 12 a$ into the quadratic formula and solve for $n$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot (- 12 a))}}{2 \cdot 1}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Raise $- 6$ to the power of $2$ .

$n = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot (- 12 a)}}{2 \cdot 1}$

Step 7.1.2

Multiply $- 4 \cdot 1 \cdot - 12$ .

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Step 7.1.2.1

Multiply $- 4$ by $1$ .

$n = \frac{6 \pm \sqrt{36 - 4 \cdot (- 12 a)}}{2 \cdot 1}$

Step 7.1.2.2

Multiply $- 4$ by $- 12$ .

$n = \frac{6 \pm \sqrt{36 + 48 a}}{2 \cdot 1}$

Step 7.1.3

Factor $12$ out of $36 + 48 a$ .

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Step 7.1.3.1

Factor $12$ out of $36$ .

$n = \frac{6 \pm \sqrt{12 (3) + 48 a}}{2 \cdot 1}$

Step 7.1.3.2

Factor $12$ out of $48 a$ .

$n = \frac{6 \pm \sqrt{12 (3) + 12 (4 a)}}{2 \cdot 1}$

Step 7.1.3.3

Factor $12$ out of $12 (3) + 12 (4 a)$ .

$n = \frac{6 \pm \sqrt{12 (3 + 4 a)}}{2 \cdot 1}$

Step 7.1.4

Rewrite $12 (3 + 4 a)$ as $2^{2} \cdot (3 (3 + 2^{2} a))$ .

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Step 7.1.4.1

Factor $4$ out of $12$ .

$n = \frac{6 \pm \sqrt{4 (3) (3 + 4 a)}}{2 \cdot 1}$

Step 7.1.4.2

Rewrite $4$ as $2^{2}$ .

$n = \frac{6 \pm \sqrt{2^{2} \cdot (3 (3 + 4 a))}}{2 \cdot 1}$

Step 7.1.4.3

Rewrite $4$ as $2^{2}$ .

$n = \frac{6 \pm \sqrt{2^{2} \cdot (3 (3 + 2^{2} a))}}{2 \cdot 1}$

Step 7.1.4.4

Add parentheses.

$n = \frac{6 \pm \sqrt{2^{2} \cdot (3 (3 + 2^{2} a))}}{2 \cdot 1}$

Step 7.1.5

Pull terms out from under the radical.

$n = \frac{6 \pm 2 \sqrt{3 (3 + 2^{2} a)}}{2 \cdot 1}$

Step 7.1.6

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

$n = \frac{6 \pm 2 \sqrt{3 (3 + 4 a)}}{2 \cdot 1}$

Step 7.2

Multiply $2$ by $1$ .

$n = \frac{6 \pm 2 \sqrt{3 (3 + 4 a)}}{2}$

Step 7.3

Simplify $\frac{6 \pm 2 \sqrt{3 (3 + 4 a)}}{2}$ .

$n = 3 \pm \sqrt{3 (3 + 4 a)}$

Step 8

The final answer is the combination of both solutions.

$n = 3 + \sqrt{3 (3 + 4 a)}$

$n = 3 - \sqrt{3 (3 + 4 a)}$