4x^2+4y^2+36y+5=0 how do i put it in standard form

4x^2+4y^2+36y+5=0
... x^2 and y^2 with equal coefficient -> circle with standard form (x-h)^2 + (y-k)^2 = r^2
... group x, y and constant
(4x^2) + (4y^2 + 36y) = -5
... eliminate x,y coefficient (div by 4)
x^2 + (y^2 + 9y) = -5/4
... add necessary term to form
x^2 + 2ax + a^2 = (x + a)^2

x^2 + [y^2 + 2(9/2)y] = -5/4
... need (9/2)^2 for y-term
x^2 + [y^2 + 2(9/2)y + (9/2)^2] = -5/4 + (9/2)^2
x^2 + (y + 9/2)^2 = 19 = (sqrt(19))^2

Answer: b 2 = 1.

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.

therefore, b 2= 1.

The answer would be 5< 6n