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