# How do you find all the real and complex roots of #r²-14r+49=0#?

##### 1 Answer

Jan 22, 2016

#### Answer:

#### Explanation:

This is a perfect square trinomial—it is equal to

#(r-7)^2=0#

This can be verified by redistributing

#(r-7)(r-7)=r^2-7r-7r+49=r^2-14r+49#

*Back to solving the equation.*

#(r-7)^2=0#

Now, we can take the square root of both sides.

#r-7=0#

#r=7#

This is the only answer. Even though quadratics typically have two answers, this has a root of