we'll do the same we did on the previous one, using the vertex form.
our vertex is at (12, -8) thus h = 12, k = -8
and our zeros are 10, 0 and 14,0, so.. .we'll use say hmm (10,0) to get the "a" coefficient.
[tex]\bf \qquad \textit{parabola vertex form}\\\\
\begin{array}{llll}
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
vertex\ (12,-8)\
\begin{cases}
h=12\\
k=-8
\end{cases}\implies y=a(x-12)^2-8
\\\\\\
\textit{we also know that }
\begin{cases}
y=0\\
x=10
\end{cases}\implies 0=a(10-12)^2-8
\\\\\\
8=a(-2)^2\implies 8=4a\implies \boxed{2=a}
\\\\\\
thus\qquad \boxed{y=2(x-12)^2-8}[/tex]
