Answer
See below
Work Step by Step
Given: $2x^2+3y^2+4x+12y-14=0$
Rewrite as: $(2x^2+4x)+(3y^2+12y)-14=0\\2(x^2+2x+1)-2+3(y^2+4y+4)-12-14=0\\2(x+1)^2+3(y+2)^2=28\\\frac{(x+1)^2}{14}+\frac{(y+2)^2}{\frac{3}{28}}=1$
Where $(h,k)=(-1,-2)$
Compare the given equation to the standard form of an equation of an ellipse.
