Answer
$-3$; $\frac{57}{13}$
Work Step by Step
Given \begin{equation}
3.25 n^2-4.5 n-42.75=0.
\end{equation} Rearrange the equation before applying the quadratic formula and determine the constants $a$, $b$, $c$: \begin{equation}
\begin{aligned}
\left(3.25 n^2-4.5 n-42.75=0\right)\cdot (4)&=0\cdot (4)\\
13 n^2-18 n-171=0&=0\\
ax^2+bx+c&=0\\
a=13,b=-18,c&=-171.
\end{aligned}
\end{equation} The quadratic formula gives: \begin{equation}
\begin{aligned}
x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\
n &=\frac{-(-18) \pm \sqrt{(-18)^2-4 \cdot 13 \cdot (-171)}}{2 \cdot 13}\\
&=\frac{18 \pm \sqrt{9216} }{26}\\
&=\frac{18\pm 96}{26}\\
&=\frac{9\pm 48}{13}.
\end{aligned}
\end{equation} The solutions are: \begin{aligned}
n &=\frac{9-48}{13}=-3\\
n&=\frac{9+48}{13}=\frac{57}{13}.
\end{aligned}
