Question 475095
How do I put this equation in standard hyperbola form AND then find the coordinates of the foci? My biggest problem is putting it in standard hyperbola form because I am bad a factoring! 
-16x^2 + 25y^2 - 32x - 250y + 209 = 0
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-16x^2+25y^2-32x-250y+209=0
completing the squares
-16(x^2+2x+1)+25(y^2-10y+25)+209+16-625=0
-16(x+1)^2+25(y-5)^2-400=0
-16(x+1)^2+25(y-5)^2=400
Divide by 400
-(x+1)^2/25+(y-5)^2/16=1
rearrange terms
(y-5)^2/16-(x+1)^2/25=1
This is a hyperbola with a vertical transverse axis of the standard form: (y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center
..
center: (-1,5)
a^2=16
a=4
length of transverse axis=2a=8
b^2=25
b=5
Length of conjugate axis=2b=10
c^2=a^2+b^2=16+25=41
c=√41=6.4
Foci: (-1,5±√41)=(-1,5±6.4)=(-1,11.4) and (-1,-1.4)