Solve (3n-1)^2=12n+8

$$( 3 n - 1 ) ^ { 2 } = 12 n + 8$$

Solve for n

$$n=-\frac{1}{3}\approx -0.333333333$$

$$n = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

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}.

$$9n^{2}-6n+1=12n+8$$

Subtract 12n from both sides.

$$9n^{2}-6n+1-12n=8$$

Combine -6n and -12n to get -18n.

$$9n^{2}-18n+1=8$$

Subtract 8 from both sides.

$$9n^{2}-18n+1-8=0$$

Subtract 8 from 1 to get -7.

$$9n^{2}-18n-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9n^{2}+an+bn-7. To find a and b, set up a system to be solved.

$$a+b=-18 ab=9\left(-7\right)=-63$$

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 -63.

$$1,-63 3,-21 7,-9$$

Calculate the sum for each pair.

$$1-63=-62 3-21=-18 7-9=-2$$

The solution is the pair that gives sum -18.

$$a=-21 b=3$$

Rewrite 9n^{2}-18n-7 as \left(9n^{2}-21n\right)+\left(3n-7\right).

$$\left(9n^{2}-21n\right)+\left(3n-7\right)$$

Factor out 3n in 9n^{2}-21n.

$$3n\left(3n-7\right)+3n-7$$

Factor out common term 3n-7 by using distributive property.

$$\left(3n-7\right)\left(3n+1\right)$$

To find equation solutions, solve 3n-7=0 and 3n+1=0.

$$n=\frac{7}{3} n=-\frac{1}{3}$$

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}.

$$9n^{2}-6n+1=12n+8$$

Subtract 12n from both sides.

$$9n^{2}-6n+1-12n=8$$

Combine -6n and -12n to get -18n.

$$9n^{2}-18n+1=8$$

Subtract 8 from both sides.

$$9n^{2}-18n+1-8=0$$

Subtract 8 from 1 to get -7.

$$9n^{2}-18n-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -18 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 9\left(-7\right)}}{2\times 9}$$

Square -18.

$$n=\frac{-\left(-18\right)±\sqrt{324-4\times 9\left(-7\right)}}{2\times 9}$$

Multiply -4 times 9.

$$n=\frac{-\left(-18\right)±\sqrt{324-36\left(-7\right)}}{2\times 9}$$

Multiply -36 times -7.

$$n=\frac{-\left(-18\right)±\sqrt{324+252}}{2\times 9}$$

Add 324 to 252.

$$n=\frac{-\left(-18\right)±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$n=\frac{-\left(-18\right)±24}{2\times 9}$$

The opposite of -18 is 18.

$$n=\frac{18±24}{2\times 9}$$

Multiply 2 times 9.

$$n=\frac{18±24}{18}$$

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

$$n=\frac{42}{18}$$

Reduce the fraction \frac{42}{18} to lowest terms by extracting and canceling out 6.

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

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

$$n=-\frac{6}{18}$$

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

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

The equation is now solved.

$$n=\frac{7}{3} n=-\frac{1}{3}$$

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}.

$$9n^{2}-6n+1=12n+8$$

Subtract 12n from both sides.

$$9n^{2}-6n+1-12n=8$$

Combine -6n and -12n to get -18n.

$$9n^{2}-18n+1=8$$

Subtract 1 from both sides.

$$9n^{2}-18n=8-1$$

Subtract 1 from 8 to get 7.

$$9n^{2}-18n=7$$

Divide both sides by 9.

$$\frac{9n^{2}-18n}{9}=\frac{7}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$n^{2}+\left(-\frac{18}{9}\right)n=\frac{7}{9}$$

Divide -18 by 9.

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

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

$$n^{2}-2n+1=\frac{7}{9}+1$$

Add \frac{7}{9} to 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

$$n=\frac{7}{3} n=-\frac{1}{3}$$