Solve by Completing the Square 3n^2-4n-6=9

$3 n^{2} - 4 n - 6 = 9$

Step 1

Move all terms not containing $n$ to the right side of the equation.

Add $6$ to both sides of the equation.

$3 n^{2} - 4 n = 9 + 6$

Add $9$ and $6$ .

$3 n^{2} - 4 n = 15$

Step 2

Divide each term in $3 n^{2} - 4 n = 15$ by $3$ and simplify.

Divide each term in $3 n^{2} - 4 n = 15$ by $3$ .

$\frac{3 n^{2}}{3} + \frac{- 4 n}{3} = \frac{15}{3}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 n^{2}}{3} + \frac{- 4 n}{3} = \frac{15}{3}$

Divide $n^{2}$ by $1$ .

$n^{2} + \frac{- 4 n}{3} = \frac{15}{3}$

Move the negative in front of the fraction.

$n^{2} - \frac{4 n}{3} = \frac{15}{3}$

Simplify the right side.

Divide $15$ by $3$ .

$n^{2} - \frac{4 n}{3} = 5$

Step 3

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{2}{3})}^{2}$

Step 4

Add the term to each side of the equation.

$n^{2} - \frac{4 n}{3} + {(- \frac{2}{3})}^{2} = 5 + {(- \frac{2}{3})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

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

Apply the product rule to $- \frac{2}{3}$ .

$n^{2} - \frac{4 n}{3} + {(- 1)}^{2} {(\frac{2}{3})}^{2} = 5 + {(- \frac{2}{3})}^{2}$

Apply the product rule to $\frac{2}{3}$ .

$n^{2} - \frac{4 n}{3} + {(- 1)}^{2} \frac{2^{2}}{3^{2}} = 5 + {(- \frac{2}{3})}^{2}$

Raise $- 1$ to the power of $2$ .

$n^{2} - \frac{4 n}{3} + 1 \frac{2^{2}}{3^{2}} = 5 + {(- \frac{2}{3})}^{2}$

Multiply $\frac{2^{2}}{3^{2}}$ by $1$ .

$n^{2} - \frac{4 n}{3} + \frac{2^{2}}{3^{2}} = 5 + {(- \frac{2}{3})}^{2}$

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

$n^{2} - \frac{4 n}{3} + \frac{4}{3^{2}} = 5 + {(- \frac{2}{3})}^{2}$

Raise $3$ to the power of $2$ .

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = 5 + {(- \frac{2}{3})}^{2}$

Simplify the right side.

Simplify $5 + {(- \frac{2}{3})}^{2}$ .

Simplify each term.

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

Apply the product rule to $- \frac{2}{3}$ .

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = 5 + {(- 1)}^{2} {(\frac{2}{3})}^{2}$

Apply the product rule to $\frac{2}{3}$ .

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = 5 + {(- 1)}^{2} \frac{2^{2}}{3^{2}}$

Raise $- 1$ to the power of $2$ .

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = 5 + 1 \frac{2^{2}}{3^{2}}$

Multiply $\frac{2^{2}}{3^{2}}$ by $1$ .

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = 5 + \frac{2^{2}}{3^{2}}$

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

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = 5 + \frac{4}{3^{2}}$

Raise $3$ to the power of $2$ .

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = 5 + \frac{4}{9}$

To write $5$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = 5 \cdot \frac{9}{9} + \frac{4}{9}$

Combine $5$ and $\frac{9}{9}$ .

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = \frac{5 \cdot 9}{9} + \frac{4}{9}$

Combine the numerators over the common denominator.

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = \frac{5 \cdot 9 + 4}{9}$

Simplify the numerator.

Multiply $5$ by $9$ .

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = \frac{45 + 4}{9}$

Add $45$ and $4$ .

$n^{2} - \frac{4 n}{3} + \frac{4}{9} = \frac{49}{9}$

Step 6

Factor the perfect trinomial square into ${(n - \frac{2}{3})}^{2}$ .

${(n - \frac{2}{3})}^{2} = \frac{49}{9}$

Step 7

Solve the equation for $n$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$n - \frac{2}{3} = \pm \sqrt{\frac{49}{9}}$

Simplify $\pm \sqrt{\frac{49}{9}}$ .

Rewrite $\sqrt{\frac{49}{9}}$ as $\frac{\sqrt{49}}{\sqrt{9}}$ .

$n - \frac{2}{3} = \pm \frac{\sqrt{49}}{\sqrt{9}}$

Simplify the numerator.

Rewrite $49$ as $7^{2}$ .

$n - \frac{2}{3} = \pm \frac{\sqrt{7^{2}}}{\sqrt{9}}$

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

$n - \frac{2}{3} = \pm \frac{7}{\sqrt{9}}$

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$n - \frac{2}{3} = \pm \frac{7}{\sqrt{3^{2}}}$

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

$n - \frac{2}{3} = \pm \frac{7}{3}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$n - \frac{2}{3} = \frac{7}{3}$

Move all terms not containing $n$ to the right side of the equation.

Add $\frac{2}{3}$ to both sides of the equation.

$n = \frac{7}{3} + \frac{2}{3}$

Combine the numerators over the common denominator.

$n = \frac{7 + 2}{3}$

Add $7$ and $2$ .

$n = \frac{9}{3}$

Divide $9$ by $3$ .

$n = 3$

Next, use the negative value of the $\pm$ to find the second solution.

$n - \frac{2}{3} = - \frac{7}{3}$

Move all terms not containing $n$ to the right side of the equation.

Add $\frac{2}{3}$ to both sides of the equation.

$n = - \frac{7}{3} + \frac{2}{3}$

Combine the numerators over the common denominator.

$n = \frac{- 7 + 2}{3}$

Add $- 7$ and $2$ .

$n = \frac{- 5}{3}$

Move the negative in front of the fraction.

$n = - \frac{5}{3}$

The complete solution is the result of both the positive and negative portions of the solution.

$n = 3, - \frac{5}{3}$

Solve by Completing the Square 3n^2-4n-6=9

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