Can some one please help me with this Write in standard form

Can some one please help me with this. Write in standard form and identify the center and the vertices. 16x^2+48x+4y^2-20y+57=0. Thank you!

Solution

16x² + 48x + 4y² - 20y + 57 = 0

regroup terms
(16x² + 48x) + (4y² - 20y) = -57

factor out leading coefficients
16(x² + 3x) + 4(y² - 5y) = -57

complete the squares
16(x² + 3x + (3/2)²) + 4(y² - 5y + (5/2)²) = -57 + 16(3/2)² + 4(5/2)²
16(x+3/2)² + 4(y+5/2)² = -57 + 36 + 25 = 4

divide both sides by the constant term, 4
4(x+3/2)² + (y+5/2)² = 1

move coefficients to denominator
(x+3/2)²/(¼) + (y+5/2)²/1 = 1
(x+3/2)²/(½)² + (y+5/2)²/1² = 1

center (-3/2, -5/2)
denominator of y term is larger than denominator of x term, so ellipse is vertical.
(y+5/2)²/1² + (x+3/2)²/(½)² = 1
vertices (-3/2, -5/2±1) = (-3/2,-3/2) and (-3/2,-7/2)
co-vertices (-3/2±½, -5/2) = (-2, -5/2) and (-1, -5/2)