Need help completing the square(math)

3x²+3x+2y=0
According to my book the answer is, y-(3/8)= (-3/2)(x+(1/2))²
How do you get this? Thanks

Move the y term to the other side.

3x^2 + 3x = -2y

Factor out the 3 from the x terms (because you cannot complete the square if there is a number with the x^2)

3(x^2 + x) = -2y

Complete the square using (b/2)^2
(1/2)^2
= 1/4
Add 1/4 to both sides of the equation(Remember to multiply it by the factored out 3 on the right sides)

3(x^2 + x + 1/4) = -2y + 3(1/4)

Factor your perfect square

3(x + 1/2)^2 = -2y + 3/4

Isolate y by dividing all terms by -2

(3(x + 1/2)^2)/-2 = y + (3/4)/-2

Simplify

(-3/2)(x + 1/2)^2 = y + (3/4) * (1/-2)
(-3/2)(x + 1/2)^2 = y + (-3/8)
(-3/2)(x + 1/2)^2 = y - 3/8

Nasty problem.

Here's some ideas:

Subtract the y-TERM to the other side.

Divide thru by 3.

That should get you (X^2 + X ) on one side.

Complete the square. Remember to add whatever to BOTH sides.

Go from there. I'll be back after lunch.

2y = -3x^2 - 3x
2y = -3 * (x^2 + x)
2y = -3 * (x^2 + 2 * (1/2) * x + (1/2)^2 - (1/2)^2)
2y = -3 * ((x + (1/2))^2 - 1/4)
2y = -3 * (x + (1/2))^2 + 3/4
2y - 3/4 = -3 * (x + (1/2))^2
y - (3/8) = (-3/2) * (x + (1/2))^2

3x² + 3x + 2y = 0

2y = -3x² - 3x

2y = -3(x² + x + _)

2y - (3/4) = -3(x² + x + (1/4))

y - (3/8) = (-3/2)(x + (1/2))²