Rewriting this in standard form we get
\begin{align*}
4x^2-5y^2-16x-50y+71 &= 4x^2-16x-5y^2-50y+71\\
&=4(x^2-4x)-5(y^2+10y)+71 \\
&= 4(x^2-4x+4)-16-5(y^2+10y+25)+125+71\\
&=4(x-2)^2-5(y+5)^2+180=0\\
&=4(x-2)^2-5(y+5)^2 = -180\\
&=5(y+5)^2-4(x-2)^2=180
\end{align*}
Dividing both sides by $180$ we get $$\frac{(y+6)^2}{36}-\frac{(x-2)^2}{45} = 1$$
This is a hyperbola with $a=6, b=3\sqrt5$ so $c = \sqrt{36+45} = 9$. The eccentricity is given by $\dfrac{c}{a}$. So it is $\dfrac96 = \dfrac32$.
Rewriting this in standard form we get \begin{align*} 4x^2-5y^2-16x-50y+71 &= 4x^2-16x-5y^2-50y+71\\ &=4(x^2-4x)-5(y^2+10y)+71 \\ &= 4(x^2-4x+4)-16-5(y^2+10y+25)+125+71\\ &=4(x-2)^2-5(y+5)^2+180=0\\ &=4(x-2)^2-5(y+5)^2 = -180\\ &=5(y+5)^2-4(x-2)^2=180 \end{align*} Dividing both sides by $180$ we get $$\frac{(y+6)^2}{36}-\frac{(x-2)^2}{45} = 1$$ This is a hyperbola with $a=6, b=3\sqrt5$ so $c = \sqrt{36+45} = 9$. The eccentricity is given by $\dfrac{c}{a}$. So it is $\dfrac96 = \dfrac32$.