How would I solve by completing the square?!:9y²-30y+23=0

9y²-30y+23=0

Thanks!

Oh jeez. Completing the square is the worst... Haha alright I'll show the steps here:

9y^2 - 30y + 23 = 0
9y^2 - 30y = -23
9(y^2 - (10/3)y) = -23
y^2 - (10/3)y = (-23/9)
y^2 - (10/3)y + _____ = (-23/9) + _____

Now IDK if you know how to do this part but ill say it anyways. You take the coefficient of y which is (10/3) and you divide it by 2 and you get (5/3) and then square that so you get (25/9) and then put that into the two blank spaces.

y^2 - (10/3)y + (25/9) = (-23/9) + (25/9) Now its perfect square trinomial
(y - (5/3))^2 = (2/9) Now you square root both sides and you get:
y - (5/3) = + or - ((sqr(2))/3)
y = (5/3) = or - ((sqr(2))/3) Now combine the two fractions
y = ((5 +or - (sqr(2)))/3) And so that is your answer.

5 +/- sqr 2
__________

3

c = (b/2)²
23 doesn't equal (30/2)^2, so put 23 on the other side of the equation:

So then the equation would be:
9y²-30y = -23

Then find c and add it to both sides
9y²-30y+15² = 15² - 23

I'll tell you that you'll end up with:
(3y-5)(3y-5) = 202

Then find the value of
of y and whatever your teacher needs you to find.

I suggest you watch this.
http://www.youtube.com/watch?v=gzm-uhj06q8
Sal Khan is a great teacher. That was how I learned to complete the square.