Solve (3n-1)^2+5=21

$$( 3 n - 1 ) ^ { 2 } + 5 = 21$$

Solve for n

$$n = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$n=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3n-1\right)^{2}.

$$9n^{2}-6n+1+5=21$$

Add 1 and 5 to get 6.

$$9n^{2}-6n+6=21$$

Subtract 21 from both sides.

$$9n^{2}-6n+6-21=0$$

Subtract 21 from 6 to get -15.

$$9n^{2}-6n-15=0$$

Divide both sides by 3.

$$3n^{2}-2n-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3n^{2}+an+bn-5. To find a and b, set up a system to be solved.

$$a+b=-2 ab=3\left(-5\right)=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite 3n^{2}-2n-5 as \left(3n^{2}-5n\right)+\left(3n-5\right).

$$\left(3n^{2}-5n\right)+\left(3n-5\right)$$

Factor out n in 3n^{2}-5n.

$$n\left(3n-5\right)+3n-5$$

Factor out common term 3n-5 by using distributive property.

$$\left(3n-5\right)\left(n+1\right)$$

To find equation solutions, solve 3n-5=0 and n+1=0.

$$n=\frac{5}{3} n=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3n-1\right)^{2}.

$$9n^{2}-6n+1+5=21$$

Add 1 and 5 to get 6.

$$9n^{2}-6n+6=21$$

Subtract 21 from both sides.

$$9n^{2}-6n+6-21=0$$

Subtract 21 from 6 to get -15.

$$9n^{2}-6n-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -6 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9\left(-15\right)}}{2\times 9}$$

Square -6.

$$n=\frac{-\left(-6\right)±\sqrt{36-4\times 9\left(-15\right)}}{2\times 9}$$

Multiply -4 times 9.

$$n=\frac{-\left(-6\right)±\sqrt{36-36\left(-15\right)}}{2\times 9}$$

Multiply -36 times -15.

$$n=\frac{-\left(-6\right)±\sqrt{36+540}}{2\times 9}$$

Add 36 to 540.

$$n=\frac{-\left(-6\right)±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$n=\frac{-\left(-6\right)±24}{2\times 9}$$

The opposite of -6 is 6.

$$n=\frac{6±24}{2\times 9}$$

Multiply 2 times 9.

$$n=\frac{6±24}{18}$$

Now solve the equation n=\frac{6±24}{18} when ± is plus. Add 6 to 24.

$$n=\frac{30}{18}$$

Reduce the fraction \frac{30}{18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation n=\frac{6±24}{18} when ± is minus. Subtract 24 from 6.

$$n=-\frac{18}{18}$$

Divide -18 by 18.

$$n=-1$$

The equation is now solved.

$$n=\frac{5}{3} n=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3n-1\right)^{2}.

$$9n^{2}-6n+1+5=21$$

Add 1 and 5 to get 6.

$$9n^{2}-6n+6=21$$

Subtract 6 from both sides.

$$9n^{2}-6n=21-6$$

Subtract 6 from 21 to get 15.

$$9n^{2}-6n=15$$

Divide both sides by 9.

$$\frac{9n^{2}-6n}{9}=\frac{15}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$n^{2}+\left(-\frac{6}{9}\right)n=\frac{15}{9}$$

Reduce the fraction \frac{-6}{9} to lowest terms by extracting and canceling out 3.

$$n^{2}-\frac{2}{3}n=\frac{15}{9}$$

Reduce the fraction \frac{15}{9} to lowest terms by extracting and canceling out 3.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{5}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{2}{3}n+\frac{1}{9}=\frac{16}{9}$$

Factor n^{2}-\frac{2}{3}n+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{1}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{1}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$n-\frac{1}{3}=\frac{4}{3} n-\frac{1}{3}=-\frac{4}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$n=\frac{5}{3} n=-1$$