Find the equation of an ellipse satisfying the given conditions.
Foci: $(-4,0)$ and $(4,0)$ : length of major axis: 14

Step 1 ：The equation of an ellipse in standard form is given by $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$, where $(h,k)$ is the center of the ellipse, $a$ is the semi-major axis length, and $b$ is the semi-minor axis length.

Step 2 ：The foci of the ellipse are located at $(h\pm c,k)$, where $c$ is the distance from the center to each focus.

Step 3 ：Given that the foci are at $(-4,0)$ and $(4,0)$, we can determine that the center of the ellipse is at the midpoint of the line segment connecting the foci, which is $(0,0)$.

Step 4 ：The distance between the foci is $2c$, so $2c=4-(-4)=8$. Therefore, $c=4$.

Step 5 ：The length of the major axis is given as 14, so $2a=14$ and $a=7$.

Step 6 ：We can find $b$ using the relationship $a^2=b^2+c^2$.

Step 7 ：Let's calculate $b$ and write the equation of the ellipse.

Step 8 ：$a=7$

Step 9 ：$c=4$

Step 10 ：$b=\sqrt{a^2-c^2}=\sqrt{49-16}=\sqrt{33}$

Step 11 ：The semi-minor axis length, $b$, is approximately 5.74. Therefore, the equation of the ellipse is $\frac{x^2}{49}+\frac{y^2}{33}=1$.

Step 12 ：Final Answer: The equation of the ellipse is $\boxed{{\displaystyle \frac{x^2}{49}+\frac{y^2}{33}=1}}$.