# How do you solve #p^ { 2} - 12p - 73= 0#?

##### 2 Answers

#### Explanation:

We can solve this by completing the square and using the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

with

#0 = p^2-12p-73#

#color(white)(0) = p^2-2p(6)+36 -36-73#

#color(white)(0) = p^2-2p(6)+36-109#

#color(white)(0) = (p-6)^2-(sqrt(109))^2#

#color(white)(0) = ((p-6)-sqrt(109))((p-6)+sqrt(109))#

#color(white)(0) = (p-6-sqrt(109))(p-6+sqrt(109))#

Hence:

#p = 6+-sqrt(109)#

#### Explanation:

Use the quadratic formula.

where

When you plug everything in, you get two answers: