Solve (6n+1)^2+3(6n+1)-4=0

$$( 6 n + 1 ) ^ { 2 } + 3 ( 6 n + 1 ) - 4 = 0$$

Solve for n

$$n=-\frac{5}{6}\approx -0.833333333$$

$$n=0$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6n+1\right)^{2}.

$$36n^{2}+12n+1+3\left(6n+1\right)-4=0$$

Use the distributive property to multiply 3 by 6n+1.

$$36n^{2}+12n+1+18n+3-4=0$$

Combine 12n and 18n to get 30n.

$$36n^{2}+30n+1+3-4=0$$

Add 1 and 3 to get 4.

$$36n^{2}+30n+4-4=0$$

Subtract 4 from 4 to get 0.

$$36n^{2}+30n=0$$

Factor out n.

$$n\left(36n+30\right)=0$$

To find equation solutions, solve n=0 and 36n+30=0.

$$n=0 n=-\frac{5}{6}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6n+1\right)^{2}.

$$36n^{2}+12n+1+3\left(6n+1\right)-4=0$$

Use the distributive property to multiply 3 by 6n+1.

$$36n^{2}+12n+1+18n+3-4=0$$

Combine 12n and 18n to get 30n.

$$36n^{2}+30n+1+3-4=0$$

Add 1 and 3 to get 4.

$$36n^{2}+30n+4-4=0$$

Subtract 4 from 4 to get 0.

$$36n^{2}+30n=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, 30 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-30±\sqrt{30^{2}}}{2\times 36}$$

Take the square root of 30^{2}.

$$n=\frac{-30±30}{2\times 36}$$

Multiply 2 times 36.

$$n=\frac{-30±30}{72}$$

Now solve the equation n=\frac{-30±30}{72} when ± is plus. Add -30 to 30.

$$n=\frac{0}{72}$$

Divide 0 by 72.

$$n=0$$

Now solve the equation n=\frac{-30±30}{72} when ± is minus. Subtract 30 from -30.

$$n=-\frac{60}{72}$$

Reduce the fraction \frac{-60}{72} to lowest terms by extracting and canceling out 12.

$$n=-\frac{5}{6}$$

The equation is now solved.

$$n=0 n=-\frac{5}{6}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6n+1\right)^{2}.

$$36n^{2}+12n+1+3\left(6n+1\right)-4=0$$

Use the distributive property to multiply 3 by 6n+1.

$$36n^{2}+12n+1+18n+3-4=0$$

Combine 12n and 18n to get 30n.

$$36n^{2}+30n+1+3-4=0$$

Add 1 and 3 to get 4.

$$36n^{2}+30n+4-4=0$$

Subtract 4 from 4 to get 0.

$$36n^{2}+30n=0$$

Divide both sides by 36.

$$\frac{36n^{2}+30n}{36}=\frac{0}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$n^{2}+\frac{30}{36}n=\frac{0}{36}$$

Reduce the fraction \frac{30}{36} to lowest terms by extracting and canceling out 6.

$$n^{2}+\frac{5}{6}n=\frac{0}{36}$$

Divide 0 by 36.

$$n^{2}+\frac{5}{6}n=0$$

Divide \frac{5}{6}, the coefficient of the x term, by 2 to get \frac{5}{12}. Then add the square of \frac{5}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{5}{6}n+\left(\frac{5}{12}\right)^{2}=\left(\frac{5}{12}\right)^{2}$$

Square \frac{5}{12} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{5}{6}n+\frac{25}{144}=\frac{25}{144}$$

Factor n^{2}+\frac{5}{6}n+\frac{25}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{5}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{5}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$n+\frac{5}{12}=\frac{5}{12} n+\frac{5}{12}=-\frac{5}{12}$$

Subtract \frac{5}{12} from both sides of the equation.

$$n=0 n=-\frac{5}{6}$$