Algebra Examples

$n \cdot 0.11 (n - 1) + 35 = 3 n$

Step 1

Simplify each term.

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Step 1.1

Rewrite using the commutative property of multiplication.

$0.11 n (n - 1) + 35 = 3 n$

Step 1.2

Apply the distributive property.

$0.11 n \cdot n + 0.11 n \cdot - 1 + 35 = 3 n$

Step 1.3

Multiply $n$ by $n$ by adding the exponents.

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Step 1.3.1

Move $n$ .

$0.11 (n \cdot n) + 0.11 n \cdot - 1 + 35 = 3 n$

Step 1.3.2

Multiply $n$ by $n$ .

$0.11 n^{2} + 0.11 n \cdot - 1 + 35 = 3 n$

Step 1.4

Multiply $- 1$ by $0.11$ .

$0.11 n^{2} - 0.11 n + 35 = 3 n$

Step 2

Move all terms containing $n$ to the left side of the equation.

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Step 2.1

Subtract $3 n$ from both sides of the equation.

$0.11 n^{2} - 0.11 n + 35 - 3 n = 0$

Step 2.2

Subtract $3 n$ from $- 0.11 n$ .

$0.11 n^{2} - 3.11 n + 35 = 0$

Step 3

Factor $0.01$ out of $0.11 n^{2} - 3.11 n + 35$ .

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Step 3.1

Factor $0.01$ out of $0.11 n^{2}$ .

$0.01 (11 n^{2}) - 3.11 n + 35 = 0$

Step 3.2

Factor $0.01$ out of $- 3.11 n$ .

$0.01 (11 n^{2}) + 0.01 (- 311 n) + 35 = 0$

Step 3.3

Factor $0.01$ out of $35$ .

$0.01 (11 n^{2}) + 0.01 (- 311 n) + 0.01 (3500) = 0$

Step 3.4

Factor $0.01$ out of $0.01 (11 n^{2}) + 0.01 (- 311 n)$ .

$0.01 (11 n^{2} - 311 n) + 0.01 (3500) = 0$

Step 3.5

Factor $0.01$ out of $0.01 (11 n^{2} - 311 n) + 0.01 (3500)$ .

$0.01 (11 n^{2} - 311 n + 3500) = 0$

Step 4

Divide each term in $0.01 (11 n^{2} - 311 n + 3500) = 0$ by $0.01$ and simplify.

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Step 4.1

Divide each term in $0.01 (11 n^{2} - 311 n + 3500) = 0$ by $0.01$ .

$\frac{0.01 (11 n^{2} - 311 n + 3500)}{0.01} = \frac{0}{0.01}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $0.01$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{0.01 (11 n^{2} - 311 n + 3500)}{0.01} = \frac{0}{0.01}$

Step 4.2.1.2

Divide $11 n^{2} - 311 n + 3500$ by $1$ .

$11 n^{2} - 311 n + 3500 = \frac{0}{0.01}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $0.01$ .

$11 n^{2} - 311 n + 3500 = 0$

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 $a = 11$ , $b = - 311$ , and $c = 3500$ into the quadratic formula and solve for $n$ .

$\frac{311 \pm \sqrt{{(- 311)}^{2} - 4 \cdot (11 \cdot 3500)}}{2 \cdot 11}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Raise $- 311$ to the power of $2$ .

$n = \frac{311 \pm \sqrt{96721 - 4 \cdot 11 \cdot 3500}}{2 \cdot 11}$

Step 7.1.2

Multiply $- 4 \cdot 11 \cdot 3500$ .

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Step 7.1.2.1

Multiply $- 4$ by $11$ .

$n = \frac{311 \pm \sqrt{96721 - 44 \cdot 3500}}{2 \cdot 11}$

Step 7.1.2.2

Multiply $- 44$ by $3500$ .

$n = \frac{311 \pm \sqrt{96721 - 154000}}{2 \cdot 11}$

Step 7.1.3

Subtract $154000$ from $96721$ .

$n = \frac{311 \pm \sqrt{- 57279}}{2 \cdot 11}$

Step 7.1.4

Rewrite $- 57279$ as $- 1 (57279)$ .

$n = \frac{311 \pm \sqrt{- 1 \cdot 57279}}{2 \cdot 11}$

Step 7.1.5

Rewrite $\sqrt{- 1 (57279)}$ as $\sqrt{- 1} \cdot \sqrt{57279}$ .

$n = \frac{311 \pm \sqrt{- 1} \cdot \sqrt{57279}}{2 \cdot 11}$

Step 7.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$n = \frac{311 \pm i \sqrt{57279}}{2 \cdot 11}$

Step 7.2

Multiply $2$ by $11$ .

$n = \frac{311 \pm i \sqrt{57279}}{22}$

Step 8

The final answer is the combination of both solutions.

$n = \frac{311 + i \sqrt{57279}}{22}, \frac{311 - i \sqrt{57279}}{22}$