Need to get into standard form.
4y^2+x^2+7y+2x-6=-3
We are given:
$\displaystyle 4y^2+x^2+7y+2x-6=-3$
The first thing I would do is add 6 to both sides:
$\displaystyle 4y^2+x^2+7y+2x=3$
Next group by like variables on the left:
$\displaystyle \left(4y^2+7y\right)+\left(x^2+2x\right)=3$
We see $\displaystyle y^2$ has a coefficient other than 1, so factor that out:
$\displaystyle 4\left(y^2+\frac{7}{4}y\right)+\left(x^2+2x\right) =3$
Now add the square of half the coefficients of the linear terms (on both sides):
$\displaystyle 4\left(y^2+\frac{7}{4}y+\frac{49}{64}\right)+\left (x^2+2x+1\right)=3+4\cdot\frac{49}{64}+1$
Rewrite:
$\displaystyle 4\left(y+\frac{7}{8}\right)^2+(x+1)^2=\frac{113}{1 6}$
Multiply through by $\displaystyle \frac{16}{113}$
$\displaystyle \frac{64}{113}\left(y+\frac{7}{8}\right)^2+\frac{1 6}{113}(x+1)^2=1$
Write in standard form:
$\displaystyle \frac{(x+1)^2}{ \left(\dfrac{\sqrt{113}}{4}\right)^2}+ \frac{\left(y+\dfrac{7}{8}\right)^2}{ \left(\dfrac{\sqrt{113}}{8}\right)^2}=1$