Solve (a+3)^2=25

$$( a + 3 ) ^ { 2 } = 25$$

Solve for a

$$a=2$$

$$a=-8$$

Steps Using Factoring

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

$$a^{2}+6a+9=25$$

Subtract 25 from both sides.

$$a^{2}+6a+9-25=0$$

Subtract 25 from 9 to get -16.

$$a^{2}+6a-16=0$$

To solve the equation, factor a^{2}+6a-16 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=6 ab=-16$$

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 -16.

$$-1,16 -2,8 -4,4$$

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 6.

$$a=-2 b=8$$

Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.

$$\left(a-2\right)\left(a+8\right)$$

To find equation solutions, solve a-2=0 and a+8=0.

$$a=2 a=-8$$

Steps Using Factoring By Grouping

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

$$a^{2}+6a+9=25$$

Subtract 25 from both sides.

$$a^{2}+6a+9-25=0$$

Subtract 25 from 9 to get -16.

$$a^{2}+6a-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba-16. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\left(-16\right)=-16$$

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 -16.

$$-1,16 -2,8 -4,4$$

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 6.

$$a=-2 b=8$$

Rewrite a^{2}+6a-16 as \left(a^{2}-2a\right)+\left(8a-16\right).

$$\left(a^{2}-2a\right)+\left(8a-16\right)$$

Factor out a in the first and 8 in the second group.

$$a\left(a-2\right)+8\left(a-2\right)$$

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

$$\left(a-2\right)\left(a+8\right)$$

To find equation solutions, solve a-2=0 and a+8=0.

$$a=2 a=-8$$

Steps Using the Quadratic Formula

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

$$a^{2}+6a+9=25$$

Subtract 25 from both sides.

$$a^{2}+6a+9-25=0$$

Subtract 25 from 9 to get -16.

$$a^{2}+6a-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-6±\sqrt{6^{2}-4\left(-16\right)}}{2}$$

Square 6.

$$a=\frac{-6±\sqrt{36-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$a=\frac{-6±\sqrt{36+64}}{2}$$

Add 36 to 64.

$$a=\frac{-6±\sqrt{100}}{2}$$

Take the square root of 100.

$$a=\frac{-6±10}{2}$$

Now solve the equation a=\frac{-6±10}{2} when ± is plus. Add -6 to 10.

$$a=\frac{4}{2}$$

Divide 4 by 2.

$$a=2$$

Now solve the equation a=\frac{-6±10}{2} when ± is minus. Subtract 10 from -6.

$$a=-\frac{16}{2}$$

Divide -16 by 2.

$$a=-8$$

The equation is now solved.

$$a=2 a=-8$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(a+3\right)^{2}}=\sqrt{25}$$

Simplify.

$$a+3=5 a+3=-5$$

Subtract 3 from both sides of the equation.

$$a=2 a=-8$$