First, group the #x#'s and the #y#'s.
#25x^2 + 200x + 9y^2 + 54y = - 480#
#=> (25x^2 + 200x) + (9y^2 + 54y) = -480#
#=> 25(x^2 + 8x) + 9(y^2 + 6y) = -480#
Then, we "complete" the square.
To do this, we add something to our binomials such that it will become a perfect square trinomial. However, when we do this, we need to add the same value to the other side of the equation so that the equality is retained
#=> 25(x^2 + 8x + 16) + 9(y^2 + 6y + 9) = -480 + 25(16) + 9(9)#
#=> 25(x + 4)^2 + 9(y + 3)^2 = -480 + 400 + 81#
#=> 25(x + 4)^2 + 9(y + 3)^2 = 1#
Since the other side of the equation is already 1, our equation is already in the form
#=> (x - h)^2/b^2 + (y - k)^2/a^2 = 1#
Now let's get the value of #a# and #b#
#=> 1/b^2(x - h)^2 + 1/a^2(y - k)^2 = 1#
#=> 1/b^2 = 25#
#=> 1/25 = b^2#
#=> b = 1/5#
#=> 1/a^2 = 9#
#=> 1/9 = a^2#
#=> a = 1/3#
Now that we have #a# and #b#, let's get #c#
#c^2 = a^2 - b^2#
#=> c^2 = (1/3)^2 - (1/5)^2#
#=> c^2 = 1/9 - 1/25#
#=> c^2 = (25 - 9)/225#
#=> c^2 = 16/225#
#=> c = 4/15#
Now that we have #c#, we can get the foci.
To get the foci, we add/subtract #c# to/from the coordinate associated with #a#. In this case, #a# is associated with #y#.
Our ellipse is centered at #(h, k) = (-4, -3)#
Our foci are at #(h, k + c), (h, k - c)#
#=> (-4, -3 + 4/15), (-4, -3 - 4/15)#
#=> (-4, -41/15), (-4, -49/15)#
