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

##### 2 Answers

#### Explanation:

First things first, in order to complete the square, the leading coefficient (

Now we have the beginning. Completing the square can look scary, but it's really just a process, and if you understand the steps, it becomes pretty simple.

The first step is getting the leading coefficient to

So, we'll take

We did all of this because we needed to find the value that will make the left side of our equation, *perfect square*, which is

Now that we have our missing value, we need to add it to our equation.

**But wait!!** We can't just add a random number into an equation! An equation is all about balance (the root word is *equal*). You can't just introduce a new value. But.... if you add *subtract*

So, if we add

Let's re-order this:

Let's simplify this:

So, now we have

Now, let's solve this puppy!

*add #10/9# to both sides*

*take a square root of both sides*

*add #1/3# to both sides*

or

#x=(sqrt10+1)/3#

#x=(-(sqrt10)+1)/3#

#### Explanation:

Given -

#3x^2-2x-3=0#

Take the constant to the right

#3x^2-2x=3#

Divide both sides by the coefficient of

#(3x^2)/3-2/3x=3/3#

#x^2-2/3x=1#

Take half of the coefficient of

Half the coefficient of

Square of

Add

#x^2-2/3x+1/9=1+1/9=(9+1)/9=10/9#

#(x-1/3)^2=10/9#

Taking square root on both sides

#x-1/3=+-sqrt (10/9)=+-sqrt10/3#

#x=+-sqrt10/3+1/3= +-(sqrt10+1)/3#

#x=(sqrt10+1)/3#

#x=(-(sqrt10)+1)/3#