Solve (2a+3)^2=16

$$( 2 a + 3 ) ^ { 2 } = 16$$

Solve for a

$$a=\frac{1}{2}=0.5$$

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

Steps Using Factoring By Grouping

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

$$4a^{2}+12a+9=16$$

Subtract 16 from both sides.

$$4a^{2}+12a+9-16=0$$

Subtract 16 from 9 to get -7.

$$4a^{2}+12a-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4a^{2}+aa+ba-7. To find a and b, set up a system to be solved.

$$a+b=12 ab=4\left(-7\right)=-28$$

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 -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 12.

$$a=-2 b=14$$

Rewrite 4a^{2}+12a-7 as \left(4a^{2}-2a\right)+\left(14a-7\right).

$$\left(4a^{2}-2a\right)+\left(14a-7\right)$$

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

$$2a\left(2a-1\right)+7\left(2a-1\right)$$

Factor out common term 2a-1 by using distributive property.

$$\left(2a-1\right)\left(2a+7\right)$$

To find equation solutions, solve 2a-1=0 and 2a+7=0.

$$a=\frac{1}{2} a=-\frac{7}{2}$$

Steps Using the Quadratic Formula

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

$$4a^{2}+12a+9=16$$

Subtract 16 from both sides.

$$4a^{2}+12a+9-16=0$$

Subtract 16 from 9 to get -7.

$$4a^{2}+12a-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 12 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-12±\sqrt{12^{2}-4\times 4\left(-7\right)}}{2\times 4}$$

Square 12.

$$a=\frac{-12±\sqrt{144-4\times 4\left(-7\right)}}{2\times 4}$$

Multiply -4 times 4.

$$a=\frac{-12±\sqrt{144-16\left(-7\right)}}{2\times 4}$$

Multiply -16 times -7.

$$a=\frac{-12±\sqrt{144+112}}{2\times 4}$$

Add 144 to 112.

$$a=\frac{-12±\sqrt{256}}{2\times 4}$$

Take the square root of 256.

$$a=\frac{-12±16}{2\times 4}$$

Multiply 2 times 4.

$$a=\frac{-12±16}{8}$$

Now solve the equation a=\frac{-12±16}{8} when ± is plus. Add -12 to 16.

$$a=\frac{4}{8}$$

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

$$a=\frac{1}{2}$$

Now solve the equation a=\frac{-12±16}{8} when ± is minus. Subtract 16 from -12.

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

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

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

The equation is now solved.

$$a=\frac{1}{2} a=-\frac{7}{2}$$

Steps for Completing the Square

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

$$4a^{2}+12a+9=16$$

Subtract 9 from both sides.

$$4a^{2}+12a=16-9$$

Subtract 9 from 16 to get 7.

$$4a^{2}+12a=7$$

Divide both sides by 4.

$$\frac{4a^{2}+12a}{4}=\frac{7}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$a^{2}+\frac{12}{4}a=\frac{7}{4}$$

Divide 12 by 4.

$$a^{2}+3a=\frac{7}{4}$$

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.

$$a^{2}+3a+\left(\frac{3}{2}\right)^{2}=\frac{7}{4}+\left(\frac{3}{2}\right)^{2}$$

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

$$a^{2}+3a+\frac{9}{4}=\frac{7+9}{4}$$

Add \frac{7}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor a^{2}+3a+\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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{3}{2}\right)^{2}}=\sqrt{4}$$

Simplify.

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

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

$$a=\frac{1}{2} a=-\frac{7}{2}$$