Answer
$\frac{(x-3)^2}{4}+y^2=1$
Work Step by Step
$x^2+4y^2-6x+5=0$
$x^2-6x+9-9+4y^2+5=0$
$(x-3)^2+4y^2=4$
$\frac{(x-3)^2}{4}+y^2=1$
We can write the general form of an ellipse:
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1$
The equation in the question is an equation of an ellipse, where $a = 2, b = 1, h = 3,$ and $k = 0$
