Basic Math Examples

${(n - 5)}^{2} = 2 n$

Step 1

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

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Step 1.1

Subtract $2 n$ from both sides of the equation.

${(n - 5)}^{2} - 2 n = 0$

Step 1.2

Simplify each term.

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Step 1.2.1

Rewrite ${(n - 5)}^{2}$ as $(n - 5) (n - 5)$ .

$(n - 5) (n - 5) - 2 n = 0$

Step 1.2.2

Expand $(n - 5) (n - 5)$ using the FOIL Method.

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Step 1.2.2.1

Apply the distributive property.

$n (n - 5) - 5 (n - 5) - 2 n = 0$

Step 1.2.2.2

Apply the distributive property.

$n \cdot n + n \cdot - 5 - 5 (n - 5) - 2 n = 0$

Step 1.2.2.3

Apply the distributive property.

$n \cdot n + n \cdot - 5 - 5 n - 5 \cdot - 5 - 2 n = 0$

Step 1.2.3

Simplify and combine like terms.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Multiply $n$ by $n$ .

$n^{2} + n \cdot - 5 - 5 n - 5 \cdot - 5 - 2 n = 0$

Step 1.2.3.1.2

Move $- 5$ to the left of $n$ .

$n^{2} - 5 \cdot n - 5 n - 5 \cdot - 5 - 2 n = 0$

Step 1.2.3.1.3

Multiply $- 5$ by $- 5$ .

$n^{2} - 5 n - 5 n + 25 - 2 n = 0$

Step 1.2.3.2

Subtract $5 n$ from $- 5 n$ .

$n^{2} - 10 n + 25 - 2 n = 0$

Step 1.3

Subtract $2 n$ from $- 10 n$ .

$n^{2} - 12 n + 25 = 0$

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 $a = 1$ , $b = - 12$ , and $c = 25$ into the quadratic formula and solve for $n$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (1 \cdot 25)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 12$ to the power of $2$ .

$n = \frac{12 \pm \sqrt{144 - 4 \cdot 1 \cdot 25}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot 25$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$n = \frac{12 \pm \sqrt{144 - 4 \cdot 25}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $25$ .

$n = \frac{12 \pm \sqrt{144 - 100}}{2 \cdot 1}$

Step 4.1.3

Subtract $100$ from $144$ .

$n = \frac{12 \pm \sqrt{44}}{2 \cdot 1}$

Step 4.1.4

Rewrite $44$ as $2^{2} \cdot 11$ .

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Step 4.1.4.1

Factor $4$ out of $44$ .

$n = \frac{12 \pm \sqrt{4 (11)}}{2 \cdot 1}$

Step 4.1.4.2

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

$n = \frac{12 \pm \sqrt{2^{2} \cdot 11}}{2 \cdot 1}$

Step 4.1.5

Pull terms out from under the radical.

$n = \frac{12 \pm 2 \sqrt{11}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$n = \frac{12 \pm 2 \sqrt{11}}{2}$

Step 4.3

Simplify $\frac{12 \pm 2 \sqrt{11}}{2}$ .

$n = 6 \pm \sqrt{11}$

Step 5

The final answer is the combination of both solutions.

$n = 6 + \sqrt{11}, 6 - \sqrt{11}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$n = 6 + \sqrt{11}, 6 - \sqrt{11}$

Decimal Form:

$n = 9.31662479 \dots, 2.68337520 \dots$