Solve (-8-c)^2=225

$$( - 8 - c ) ^ { 2 } = 225$$

Solve for c

$$c=-23$$

$$c=7$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-8-c\right)^{2}.

$$64+16c+c^{2}=225$$

Subtract 225 from both sides.

$$64+16c+c^{2}-225=0$$

Subtract 225 from 64 to get -161.

$$-161+16c+c^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$c^{2}+16c-161=0$$

To solve the equation, factor c^{2}+16c-161 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.

$$a+b=16 ab=-161$$

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 -161.

$$-1,161 -7,23$$

Calculate the sum for each pair.

$$-1+161=160 -7+23=16$$

The solution is the pair that gives sum 16.

$$a=-7 b=23$$

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

$$\left(c-7\right)\left(c+23\right)$$

To find equation solutions, solve c-7=0 and c+23=0.

$$c=7 c=-23$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-8-c\right)^{2}.

$$64+16c+c^{2}=225$$

Subtract 225 from both sides.

$$64+16c+c^{2}-225=0$$

Subtract 225 from 64 to get -161.

$$-161+16c+c^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$c^{2}+16c-161=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as c^{2}+ac+bc-161. To find a and b, set up a system to be solved.

$$a+b=16 ab=1\left(-161\right)=-161$$

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 -161.

$$-1,161 -7,23$$

Calculate the sum for each pair.

$$-1+161=160 -7+23=16$$

The solution is the pair that gives sum 16.

$$a=-7 b=23$$

Rewrite c^{2}+16c-161 as \left(c^{2}-7c\right)+\left(23c-161\right).

$$\left(c^{2}-7c\right)+\left(23c-161\right)$$

Factor out c in the first and 23 in the second group.

$$c\left(c-7\right)+23\left(c-7\right)$$

Factor out common term c-7 by using distributive property.

$$\left(c-7\right)\left(c+23\right)$$

To find equation solutions, solve c-7=0 and c+23=0.

$$c=7 c=-23$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-8-c\right)^{2}.

$$64+16c+c^{2}=225$$

Subtract 225 from both sides.

$$64+16c+c^{2}-225=0$$

Subtract 225 from 64 to get -161.

$$-161+16c+c^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$c^{2}+16c-161=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 16 for b, and -161 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-16±\sqrt{16^{2}-4\left(-161\right)}}{2}$$

Square 16.

$$c=\frac{-16±\sqrt{256-4\left(-161\right)}}{2}$$

Multiply -4 times -161.

$$c=\frac{-16±\sqrt{256+644}}{2}$$

Add 256 to 644.

$$c=\frac{-16±\sqrt{900}}{2}$$

Take the square root of 900.

$$c=\frac{-16±30}{2}$$

Now solve the equation c=\frac{-16±30}{2} when ± is plus. Add -16 to 30.

$$c=\frac{14}{2}$$

Divide 14 by 2.

$$c=7$$

Now solve the equation c=\frac{-16±30}{2} when ± is minus. Subtract 30 from -16.

$$c=-\frac{46}{2}$$

Divide -46 by 2.

$$c=-23$$

The equation is now solved.

$$c=7 c=-23$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-8-c\right)^{2}.

$$64+16c+c^{2}=225$$

Subtract 64 from both sides.

$$16c+c^{2}=225-64$$

Subtract 64 from 225 to get 161.

$$16c+c^{2}=161$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$c^{2}+16c=161$$

Divide 16, the coefficient of the x term, by 2 to get 8. Then add the square of 8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}+16c+8^{2}=161+8^{2}$$

Square 8.

$$c^{2}+16c+64=161+64$$

Add 161 to 64.

$$c^{2}+16c+64=225$$

Factor c^{2}+16c+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(c+8\right)^{2}=225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c+8\right)^{2}}=\sqrt{225}$$

Simplify.

$$c+8=15 c+8=-15$$

Subtract 8 from both sides of the equation.

$$c=7 c=-23$$