standard form of the ellipse help

$4x^2+ 3y^2 - 16x + 9y + 16 = 0\\ 4(x^2 - 4x) + 3(y^2 + 3y) + 16 = 0\\ 4(x^2 - 4x + 4 - 4) + 3\left(y^2 + 3y+\dfrac 9 4 - \dfrac 9 4\right) + 16 = 0\\ 4(x-2)^2 - 16 + 3\left(y+\dfrac 3 2\right)^2- \dfrac{27}{4}+16 = 0\\ \left(\dfrac{x-2}{\frac 1 2}\right)^2 + \left(\dfrac{y+\frac 3 2}{\frac{1}{\sqrt{3}}}\right)^2 = \left(\dfrac{3\sqrt{3}}{2}\right)^2$