Solve (2x+3)^2=36

$$( 2 x + 3 ) ^ { 2 } = 36$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x = -\frac{9}{2} = -4\frac{1}{2} = -4.5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}+12x+9=36$$

Subtract 36 from both sides.

$$4x^{2}+12x+9-36=0$$

Subtract 36 from 9 to get -27.

$$4x^{2}+12x-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=12 ab=4\left(-27\right)=-108$$

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 -108.

$$-1,108 -2,54 -3,36 -4,27 -6,18 -9,12$$

Calculate the sum for each pair.

$$-1+108=107 -2+54=52 -3+36=33 -4+27=23 -6+18=12 -9+12=3$$

The solution is the pair that gives sum 12.

$$a=-6 b=18$$

Rewrite 4x^{2}+12x-27 as \left(4x^{2}-6x\right)+\left(18x-27\right).

$$\left(4x^{2}-6x\right)+\left(18x-27\right)$$

Factor out 2x in the first and 9 in the second group.

$$2x\left(2x-3\right)+9\left(2x-3\right)$$

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

$$\left(2x-3\right)\left(2x+9\right)$$

To find equation solutions, solve 2x-3=0 and 2x+9=0.

$$x=\frac{3}{2} x=-\frac{9}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}+12x+9=36$$

Subtract 36 from both sides.

$$4x^{2}+12x+9-36=0$$

Subtract 36 from 9 to get -27.

$$4x^{2}+12x-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 12 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 4\left(-27\right)}}{2\times 4}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 4\left(-27\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-12±\sqrt{144-16\left(-27\right)}}{2\times 4}$$

Multiply -16 times -27.

$$x=\frac{-12±\sqrt{144+432}}{2\times 4}$$

Add 144 to 432.

$$x=\frac{-12±\sqrt{576}}{2\times 4}$$

Take the square root of 576.

$$x=\frac{-12±24}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-12±24}{8}$$

Now solve the equation x=\frac{-12±24}{8} when ± is plus. Add -12 to 24.

$$x=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-12±24}{8} when ± is minus. Subtract 24 from -12.

$$x=-\frac{36}{8}$$

Reduce the fraction \frac{-36}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{9}{2}$$

The equation is now solved.

$$x=\frac{3}{2} x=-\frac{9}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}+12x+9=36$$

Subtract 9 from both sides.

$$4x^{2}+12x=36-9$$

Subtract 9 from 36 to get 27.

$$4x^{2}+12x=27$$

Divide both sides by 4.

$$\frac{4x^{2}+12x}{4}=\frac{27}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{12}{4}x=\frac{27}{4}$$

Divide 12 by 4.

$$x^{2}+3x=\frac{27}{4}$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=\frac{27}{4}+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=\frac{27+9}{4}$$

Add \frac{27}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=9$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{9}$$

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{3}{2} x=-\frac{9}{2}$$