math
posted by Bolu .
The foci and vertice of 2x^2+5y^2+8x+10y+3=0

let's complete the square
2(x^2 + 4x + ....) + 5(y^2 + 2y + ...) = 3
2(x^2 + 4x + 4) + 5(y^2 + 2y + 1) = 3 + 8 + 5
2(x+2)^2 + 5(y+1)^2 = 10
divide by 10
(x+2)^2/5 + (y+1)^2/2 = 1
can you take it from there?
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