Answer
$e=\dfrac{\sqrt{21}}{7} \approx 0.654653$
Work Step by Step
Divide both sides of the equation by $28$ to obtain:
$$\dfrac{x^2}{7}+\dfrac{y^2}{4}=1$$
The ellipse above has $a^2=7$ and $b^2=4$.
Solve for $c$ using the formula $c^2=a^2-b^2$ to obtain:
\begin{align*}
c^2&=a^2-b^2\\
c^2&=7-4\\
c^2&=3\\
c&=\sqrt3 \quad\text{(since c is non-negative, we only take the principal root)}
\end{align*}
The eccentricity of the ellipse can be computed as: $e=\dfrac{c}{a}$
Solve for $a$:
$$a^2=7\longrightarrow a=\sqrt7$$
Thus,
$$e=\dfrac{\sqrt3}{\sqrt7}=\dfrac{\sqrt3}{\sqrt7}\cdot \dfrac{\sqrt7}{\sqrt7}=\dfrac{\sqrt{21}}{7} \approx 0.65$$
