To solve #2x-3x^2=-8# we will collect terms on one side. I prefer a positive in front of #x^2#, so I'll collect them to make that happen:

#2x-3x^2=-8#

#0 = 3x^2-2x-8#

Now, of course, **this** equals **that** exactly when **that** equals **this**, so we can write:

#3x^2 - 2x -8 =0#

There are a couple of ways to fill in the details of completing the square, but discussing all possibilities is more confusing than helpful, so we'll go through it one way.

#3x^2 - 2x -8 =0#

On the left, we want a square like #(x-a)^2 = x^2 - 2ax + a^2#. Let's get that #8# out of our way:

#3x^2 - 2x = 8#

We don't want that #3# out front, so we'll multiply both sides by #1/3# (Don't forget to distribute on the left.)

#1/3(3x^2 - 2x) =1/3 (8)#

#x^2 - 2/3 x = 8/3#

Cookbook: now take #1/2# of the number in front of #x#, square that and then add that square to both sides. (Why later, in the Note below.)

#1/2 * 2/3 = 1/3#

#(1/3)^2 = 1/9#

Add #1/9# to both sides:

#x^2 - 2/3 x +1/9= 8/3+1/9#

Now we can factor on the left -- remember the #1/3# we squared? That's what we need now:

#(x-1/3)^2 = 24/9 +1/9#

Notice that I also got a common denominator on the right so I can do the addition on the right:

#(x-1/3)^2 = 25/9 #

Now #a^2 = n# when #a = # either #sqrtn# or #-sqrtn#. So

#x-1/3 = +- sqrt(25/9)# Simplify the right, to get

#x-1/3 = +- 5/3# Add #1/3# to both sides:

#x= 1/3 +- 5/3#

Remember that this means there are two solutions.

One of them is #1/3 + 5/3 = 6/3 =2#

and the other is #1/3 - 5/3 = (-4)/3 = -4/3#

**Note**

When we got

#x^2 - 2/3 x = 8/3# Why did we do what we did?

We want

#x^2 - 2/3 x + "something"# to be a perfect square like:

#x^2 -2ax +a^2#

So the number in front of #x# is 2 times the thing I need to see the square of. That is #2/3 = 2a#. To fins #a#, take #1/2# of the number in front of #x#.