Solve left(6x+1right)^2=25

$${ \left(6x+1 \right) }^{ 2 } = 25$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=-1$$

Steps Using Factoring By Grouping

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

$$36x^{2}+12x+1=25$$

Subtract 25 from both sides.

$$36x^{2}+12x+1-25=0$$

Subtract 25 from 1 to get -24.

$$36x^{2}+12x-24=0$$

Divide both sides by 12.

$$3x^{2}+x-2=0$$

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

$$a+b=1 ab=3\left(-2\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

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

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

Factor out x in 3x^{2}-2x.

$$x\left(3x-2\right)+3x-2$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(x+1\right)$$

To find equation solutions, solve 3x-2=0 and x+1=0.

$$x=\frac{2}{3} x=-1$$

Steps Using the Quadratic Formula

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

$$36x^{2}+12x+1=25$$

Subtract 25 from both sides.

$$36x^{2}+12x+1-25=0$$

Subtract 25 from 1 to get -24.

$$36x^{2}+12x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, 12 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 36\left(-24\right)}}{2\times 36}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 36\left(-24\right)}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-12±\sqrt{144-144\left(-24\right)}}{2\times 36}$$

Multiply -144 times -24.

$$x=\frac{-12±\sqrt{144+3456}}{2\times 36}$$

Add 144 to 3456.

$$x=\frac{-12±\sqrt{3600}}{2\times 36}$$

Take the square root of 3600.

$$x=\frac{-12±60}{2\times 36}$$

Multiply 2 times 36.

$$x=\frac{-12±60}{72}$$

Now solve the equation x=\frac{-12±60}{72} when ± is plus. Add -12 to 60.

$$x=\frac{48}{72}$$

Reduce the fraction \frac{48}{72} to lowest terms by extracting and canceling out 24.

$$x=\frac{2}{3}$$

Now solve the equation x=\frac{-12±60}{72} when ± is minus. Subtract 60 from -12.

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

Divide -72 by 72.

$$x=-1$$

The equation is now solved.

$$x=\frac{2}{3} x=-1$$

Steps for Completing the Square

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

$$36x^{2}+12x+1=25$$

Subtract 1 from both sides.

$$36x^{2}+12x=25-1$$

Subtract 1 from 25 to get 24.

$$36x^{2}+12x=24$$

Divide both sides by 36.

$$\frac{36x^{2}+12x}{36}=\frac{24}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$x^{2}+\frac{12}{36}x=\frac{24}{36}$$

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

$$x^{2}+\frac{1}{3}x=\frac{24}{36}$$

Reduce the fraction \frac{24}{36} to lowest terms by extracting and canceling out 12.

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

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.

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

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

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

Add \frac{2}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{25}{36}$$

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}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{5}{6} x+\frac{1}{6}=-\frac{5}{6}$$

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

$$x=\frac{2}{3} x=-1$$