hmmm so.. .it could be either a horizontal or vertical parabola, let's assume or use a vertical parabola vertex equation then... namely, the squared variable being "x".
[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}})\\\\
-------------------------------\\\\
\begin{cases}
h=6\\
k=-5
\end{cases}\implies y=a(x-6)^2-5
\\\\\\
\textit{we also know }
\begin{cases}
x=0\\
y=175
\end{cases}\implies 175=a(0-6)^2-5\implies 180=a6^2
\\\\\\
\cfrac{180}{36}=a\implies 5=a\qquad thus\qquad \boxed{y=5(x-6)^2-5}[/tex]
[tex]\bf \textit{at the x-intercepts, y = 0, thus }\qquad 0=5(x-6)^2-5
\\\\\\
5=5(x-6)^2\implies \cfrac{5}{5}=(x-6)^2\implies \pm\sqrt{1}=x-6
\\\\\\
\pm 1+6=x\implies x=
\begin{cases}
7\\
5
\end{cases}[/tex]
