Question 688050
 put the equation in standard form? The vertices and the center of the equation are also needed: 
-16x^2-4y^2 = 48x-20y+57 
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Rearrange:
16x^2+48x + 4y^2 -20y + 57 = 0 
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16(x^2+3x) + 4(y^2-5y) + 57 = 0
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Complete the square:
16(x^2 + 3x + (3/2)^2) + 4(y^2 - 5y + (5/2)^2) = -57 + 16(3/2)^2 + 4(5/2)^2
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16(x+(3/2))^2 + 4(y+(5/2))^2 = -57 + 36 + 25
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16(x+(3/2))^2/(56/16) + 4(y+(5/2)^2/(56/4) = 4
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(x+(3/2))^2/(1/4) + (y+(5/2))^2/1 = 1
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Center at (-3/2 , -5/2)
Vertices at (-3/2, (-5/2)-1) and (-3/2, (-5/2)+1)
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Cheers,
Stan H.
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