Solve (c+9)^2=64

$$( c + 9 ) ^ { 2 } = 64$$

Solve for c

$$c=-1$$

$$c=-17$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(c+9\right)^{2}.

$$c^{2}+18c+81=64$$

Subtract 64 from both sides.

$$c^{2}+18c+81-64=0$$

Subtract 64 from 81 to get 17.

$$c^{2}+18c+17=0$$

To solve the equation, factor c^{2}+18c+17 using formula c^{2}+\left(a+b\right)c+ab=\left(c+a\right)\left(c+b\right). To find a and b, set up a system to be solved.

$$a+b=18 ab=17$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=17$$

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

$$\left(c+1\right)\left(c+17\right)$$

To find equation solutions, solve c+1=0 and c+17=0.

$$c=-1 c=-17$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(c+9\right)^{2}.

$$c^{2}+18c+81=64$$

Subtract 64 from both sides.

$$c^{2}+18c+81-64=0$$

Subtract 64 from 81 to get 17.

$$c^{2}+18c+17=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as c^{2}+ac+bc+17. To find a and b, set up a system to be solved.

$$a+b=18 ab=1\times 17=17$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=17$$

Rewrite c^{2}+18c+17 as \left(c^{2}+c\right)+\left(17c+17\right).

$$\left(c^{2}+c\right)+\left(17c+17\right)$$

Factor out c in the first and 17 in the second group.

$$c\left(c+1\right)+17\left(c+1\right)$$

Factor out common term c+1 by using distributive property.

$$\left(c+1\right)\left(c+17\right)$$

To find equation solutions, solve c+1=0 and c+17=0.

$$c=-1 c=-17$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(c+9\right)^{2}.

$$c^{2}+18c+81=64$$

Subtract 64 from both sides.

$$c^{2}+18c+81-64=0$$

Subtract 64 from 81 to get 17.

$$c^{2}+18c+17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 18 for b, and 17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-18±\sqrt{18^{2}-4\times 17}}{2}$$

Square 18.

$$c=\frac{-18±\sqrt{324-4\times 17}}{2}$$

Multiply -4 times 17.

$$c=\frac{-18±\sqrt{324-68}}{2}$$

Add 324 to -68.

$$c=\frac{-18±\sqrt{256}}{2}$$

Take the square root of 256.

$$c=\frac{-18±16}{2}$$

Now solve the equation c=\frac{-18±16}{2} when ± is plus. Add -18 to 16.

$$c=-\frac{2}{2}$$

Divide -2 by 2.

$$c=-1$$

Now solve the equation c=\frac{-18±16}{2} when ± is minus. Subtract 16 from -18.

$$c=-\frac{34}{2}$$

Divide -34 by 2.

$$c=-17$$

The equation is now solved.

$$c=-1 c=-17$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(c+9\right)^{2}}=\sqrt{64}$$

Simplify.

$$c+9=8 c+9=-8$$

Subtract 9 from both sides of the equation.

$$c=-1 c=-17$$