Solve (p-6)^2=100

$$( p - 6 ) ^ { 2 } = 100$$

Solve for p

$$p=16$$

$$p=-4$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(p-6\right)^{2}.

$$p^{2}-12p+36=100$$

Subtract 100 from both sides.

$$p^{2}-12p+36-100=0$$

Subtract 100 from 36 to get -64.

$$p^{2}-12p-64=0$$

To solve the equation, factor p^{2}-12p-64 using formula p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right). To find a and b, set up a system to be solved.

$$a+b=-12 ab=-64$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -64.

$$1,-64 2,-32 4,-16 8,-8$$

Calculate the sum for each pair.

$$1-64=-63 2-32=-30 4-16=-12 8-8=0$$

The solution is the pair that gives sum -12.

$$a=-16 b=4$$

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

$$\left(p-16\right)\left(p+4\right)$$

To find equation solutions, solve p-16=0 and p+4=0.

$$p=16 p=-4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(p-6\right)^{2}.

$$p^{2}-12p+36=100$$

Subtract 100 from both sides.

$$p^{2}-12p+36-100=0$$

Subtract 100 from 36 to get -64.

$$p^{2}-12p-64=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp-64. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\left(-64\right)=-64$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -64.

$$1,-64 2,-32 4,-16 8,-8$$

Calculate the sum for each pair.

$$1-64=-63 2-32=-30 4-16=-12 8-8=0$$

The solution is the pair that gives sum -12.

$$a=-16 b=4$$

Rewrite p^{2}-12p-64 as \left(p^{2}-16p\right)+\left(4p-64\right).

$$\left(p^{2}-16p\right)+\left(4p-64\right)$$

Factor out p in the first and 4 in the second group.

$$p\left(p-16\right)+4\left(p-16\right)$$

Factor out common term p-16 by using distributive property.

$$\left(p-16\right)\left(p+4\right)$$

To find equation solutions, solve p-16=0 and p+4=0.

$$p=16 p=-4$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(p-6\right)^{2}.

$$p^{2}-12p+36=100$$

Subtract 100 from both sides.

$$p^{2}-12p+36-100=0$$

Subtract 100 from 36 to get -64.

$$p^{2}-12p-64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-64\right)}}{2}$$

Square -12.

$$p=\frac{-\left(-12\right)±\sqrt{144-4\left(-64\right)}}{2}$$

Multiply -4 times -64.

$$p=\frac{-\left(-12\right)±\sqrt{144+256}}{2}$$

Add 144 to 256.

$$p=\frac{-\left(-12\right)±\sqrt{400}}{2}$$

Take the square root of 400.

$$p=\frac{-\left(-12\right)±20}{2}$$

The opposite of -12 is 12.

$$p=\frac{12±20}{2}$$

Now solve the equation p=\frac{12±20}{2} when ± is plus. Add 12 to 20.

$$p=\frac{32}{2}$$

Divide 32 by 2.

$$p=16$$

Now solve the equation p=\frac{12±20}{2} when ± is minus. Subtract 20 from 12.

$$p=-\frac{8}{2}$$

Divide -8 by 2.

$$p=-4$$

The equation is now solved.

$$p=16 p=-4$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(p-6\right)^{2}}=\sqrt{100}$$

Simplify.

$$p-6=10 p-6=-10$$

Add 6 to both sides of the equation.

$$p=16 p=-4$$