Solve (6d+1)^2+12=13

$$( 6 d + 1 ) ^ { 2 } + 12 = 13$$

Solve for d

$$d=0$$

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

Steps Using Factoring

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

$$36d^{2}+12d+1+12=13$$

Add 1 and 12 to get 13.

$$36d^{2}+12d+13=13$$

Subtract 13 from both sides.

$$36d^{2}+12d+13-13=0$$

Subtract 13 from 13 to get 0.

$$36d^{2}+12d=0$$

Factor out d.

$$d\left(36d+12\right)=0$$

To find equation solutions, solve d=0 and 36d+12=0.

$$d=0 d=-\frac{1}{3}$$

Steps Using the Quadratic Formula

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

$$36d^{2}+12d+1+12=13$$

Add 1 and 12 to get 13.

$$36d^{2}+12d+13=13$$

Subtract 13 from both sides.

$$36d^{2}+12d+13-13=0$$

Subtract 13 from 13 to get 0.

$$36d^{2}+12d=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, 12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$d=\frac{-12±\sqrt{12^{2}}}{2\times 36}$$

Take the square root of 12^{2}.

$$d=\frac{-12±12}{2\times 36}$$

Multiply 2 times 36.

$$d=\frac{-12±12}{72}$$

Now solve the equation d=\frac{-12±12}{72} when ± is plus. Add -12 to 12.

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

Divide 0 by 72.

$$d=0$$

Now solve the equation d=\frac{-12±12}{72} when ± is minus. Subtract 12 from -12.

$$d=-\frac{24}{72}$$

Reduce the fraction \frac{-24}{72} to lowest terms by extracting and canceling out 24.

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

The equation is now solved.

$$d=0 d=-\frac{1}{3}$$

Steps for Completing the Square

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

$$36d^{2}+12d+1+12=13$$

Add 1 and 12 to get 13.

$$36d^{2}+12d+13=13$$

Subtract 13 from both sides.

$$36d^{2}+12d=13-13$$

Subtract 13 from 13 to get 0.

$$36d^{2}+12d=0$$

Divide both sides by 36.

$$\frac{36d^{2}+12d}{36}=\frac{0}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$d^{2}+\frac{12}{36}d=\frac{0}{36}$$

Reduce the fraction \frac{12}{36} to lowest terms by extracting and canceling out 12.

$$d^{2}+\frac{1}{3}d=\frac{0}{36}$$

Divide 0 by 36.

$$d^{2}+\frac{1}{3}d=0$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$d^{2}+\frac{1}{3}d+\left(\frac{1}{6}\right)^{2}=\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$d^{2}+\frac{1}{3}d+\frac{1}{36}=\frac{1}{36}$$

Factor d^{2}+\frac{1}{3}d+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(d+\frac{1}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(d+\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$d+\frac{1}{6}=\frac{1}{6} d+\frac{1}{6}=-\frac{1}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$d=0 d=-\frac{1}{3}$$