# How do you solve #y^2-5y-3=0# by completing the square?

##### 2 Answers

#### Explanation:

We will also use the difference of squares identity, which can be written:

#a^2-b^2=(a-b)(a+b)#

with

Pre-multiply by

#0 = 4(y^2-5y-3)#

#=4y^2-20y-12#

#=(2y-5)^2-25-12#

#=(2y-5)^2-(sqrt(37))^2#

#=((2y-5)-sqrt(37))((2y-5)+sqrt(37))#

#=(2y-5-sqrt(37))(2y-5+sqrt(37))#

Hence:

#y = 5/2+-sqrt(37)/2#

#### Explanation:

Complete the square by adding what is missing from the square of the binomial to BOTH sides.