let's firstly find the equation of the parabola, bearing in mind that x-intercepts or solutions/zeros/roots means y = 0.
[tex]\bf ~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=1\\ k=-9 \end{cases}\implies y=a(x-1)^2-9 \\\\\\ \textit{we also know that } \begin{cases} x=0\\ y=-6 \end{cases}\implies -6=a(0-1)^2-9\implies 3=a(-1)^2[/tex]
[tex]\bf 3=a\qquad \qquad \textit{therefore}\qquad \qquad \boxed{y=3(x-1)^2-9} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{y}{0}=3(x-1)^2-9\implies 9=3(x-1)^2\implies \cfrac{9}{3}=(x-1)^2\implies 3=(x-1)^2 \\\\\\ \pm\sqrt{3}=x-1\implies \pm\sqrt{3}+1=x\implies x= \begin{cases} \sqrt{3}+1\\ -\sqrt{3}+1 \end{cases}\implies x\approx \begin{cases} 2.73\\ -0.73 \end{cases}[/tex]
