Solve 3(2x+1)^2=108

$$3 ( 2 x + 1 ) ^ { 2 } = 108$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

Steps Using Factoring By Grouping

Divide both sides by 3.

$$\left(2x+1\right)^{2}=\frac{108}{3}$$

Divide 108 by 3 to get 36.

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

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

$$4x^{2}+4x+1=36$$

Subtract 36 from both sides.

$$4x^{2}+4x+1-36=0$$

Subtract 36 from 1 to get -35.

$$4x^{2}+4x-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=4 ab=4\left(-35\right)=-140$$

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 -140.

$$-1,140 -2,70 -4,35 -5,28 -7,20 -10,14$$

Calculate the sum for each pair.

$$-1+140=139 -2+70=68 -4+35=31 -5+28=23 -7+20=13 -10+14=4$$

The solution is the pair that gives sum 4.

$$a=-10 b=14$$

Rewrite 4x^{2}+4x-35 as \left(4x^{2}-10x\right)+\left(14x-35\right).

$$\left(4x^{2}-10x\right)+\left(14x-35\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(2x-5\right)+7\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(2x+7\right)$$

To find equation solutions, solve 2x-5=0 and 2x+7=0.

$$x=\frac{5}{2} x=-\frac{7}{2}$$

Steps Using the Quadratic Formula

Divide both sides by 3.

$$\left(2x+1\right)^{2}=\frac{108}{3}$$

Divide 108 by 3 to get 36.

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

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

$$4x^{2}+4x+1=36$$

Subtract 36 from both sides.

$$4x^{2}+4x+1-36=0$$

Subtract 36 from 1 to get -35.

$$4x^{2}+4x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 4 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\left(-35\right)}}{2\times 4}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 4\left(-35\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-4±\sqrt{16-16\left(-35\right)}}{2\times 4}$$

Multiply -16 times -35.

$$x=\frac{-4±\sqrt{16+560}}{2\times 4}$$

Add 16 to 560.

$$x=\frac{-4±\sqrt{576}}{2\times 4}$$

Take the square root of 576.

$$x=\frac{-4±24}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-4±24}{8}$$

Now solve the equation x=\frac{-4±24}{8} when ± is plus. Add -4 to 24.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{-4±24}{8} when ± is minus. Subtract 24 from -4.

$$x=-\frac{28}{8}$$

Reduce the fraction \frac{-28}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{7}{2}$$

The equation is now solved.

$$x=\frac{5}{2} x=-\frac{7}{2}$$

Steps for Completing the Square

Divide both sides by 3.

$$\left(2x+1\right)^{2}=\frac{108}{3}$$

Divide 108 by 3 to get 36.

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

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

$$4x^{2}+4x+1=36$$

Subtract 1 from both sides.

$$4x^{2}+4x=36-1$$

Subtract 1 from 36 to get 35.

$$4x^{2}+4x=35$$

Divide both sides by 4.

$$\frac{4x^{2}+4x}{4}=\frac{35}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{4}{4}x=\frac{35}{4}$$

Divide 4 by 4.

$$x^{2}+x=\frac{35}{4}$$

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{35}{4}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{35+1}{4}$$

Add \frac{35}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=9$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{9}$$

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{5}{2} x=-\frac{7}{2}$$