Solve (6x+7)^2-12(6x+7)+36=0

$$( 6 x + 7 ) ^ { 2 } - 12 ( 6 x + 7 ) + 36 = 0$$

Solve for x

$$x=-\frac{1}{6}\approx -0.166666667$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6x+7\right)^{2}.

$$36x^{2}+84x+49-12\left(6x+7\right)+36=0$$

Use the distributive property to multiply -12 by 6x+7.

$$36x^{2}+84x+49-72x-84+36=0$$

Combine 84x and -72x to get 12x.

$$36x^{2}+12x+49-84+36=0$$

Subtract 84 from 49 to get -35.

$$36x^{2}+12x-35+36=0$$

Add -35 and 36 to get 1.

$$36x^{2}+12x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 36x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=12 ab=36\times 1=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 12.

$$a=6 b=6$$

Rewrite 36x^{2}+12x+1 as \left(36x^{2}+6x\right)+\left(6x+1\right).

$$\left(36x^{2}+6x\right)+\left(6x+1\right)$$

Factor out 6x in 36x^{2}+6x.

$$6x\left(6x+1\right)+6x+1$$

Factor out common term 6x+1 by using distributive property.

$$\left(6x+1\right)\left(6x+1\right)$$

Rewrite as a binomial square.

$$\left(6x+1\right)^{2}$$

To find equation solution, solve 6x+1=0.

$$x=-\frac{1}{6}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6x+7\right)^{2}.

$$36x^{2}+84x+49-12\left(6x+7\right)+36=0$$

Use the distributive property to multiply -12 by 6x+7.

$$36x^{2}+84x+49-72x-84+36=0$$

Combine 84x and -72x to get 12x.

$$36x^{2}+12x+49-84+36=0$$

Subtract 84 from 49 to get -35.

$$36x^{2}+12x-35+36=0$$

Add -35 and 36 to get 1.

$$36x^{2}+12x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, 12 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 36}}{2\times 36}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 36}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-12±\sqrt{144-144}}{2\times 36}$$

Add 144 to -144.

$$x=\frac{-12±\sqrt{0}}{2\times 36}$$

Take the square root of 0.

$$x=-\frac{12}{2\times 36}$$

Multiply 2 times 36.

$$x=-\frac{12}{72}$$

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

$$x=-\frac{1}{6}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6x+7\right)^{2}.

$$36x^{2}+84x+49-12\left(6x+7\right)+36=0$$

Use the distributive property to multiply -12 by 6x+7.

$$36x^{2}+84x+49-72x-84+36=0$$

Combine 84x and -72x to get 12x.

$$36x^{2}+12x+49-84+36=0$$

Subtract 84 from 49 to get -35.

$$36x^{2}+12x-35+36=0$$

Add -35 and 36 to get 1.

$$36x^{2}+12x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$36x^{2}+12x=-1$$

Divide both sides by 36.

$$\frac{36x^{2}+12x}{36}=-\frac{1}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$x^{2}+\frac{12}{36}x=-\frac{1}{36}$$

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

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

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.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=-\frac{1}{36}+\left(\frac{1}{6}\right)^{2}$$

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

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

Add -\frac{1}{36} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}+\frac{1}{3}x+\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}.

$$\left(x+\frac{1}{6}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+\frac{1}{6}=0 x+\frac{1}{6}=0$$

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

$$x=-\frac{1}{6} x=-\frac{1}{6}$$

The equation is now solved. Solutions are the same.

$$x=-\frac{1}{6}$$