[tex]\begin{cases} x = 3+\sqrt{5}\implies &x-3-\sqrt{5}=0\\ x= 3-\sqrt{5}\implies &x-3+\sqrt{5}=0 \end{cases}\implies a(x-3-\sqrt{5})(x-3+\sqrt{5})=\stackrel{y}{0} \\\\[-0.35em] ~\dotfill\\\\ \underset{\textit{difference of squares}}{[(x-3)~-~(\sqrt{5})][(x-3)~+~(\sqrt{5})]}\implies (x-3)^2-(\sqrt{5})^2 \\\\\\ (x^2-6x+9)-5\implies x^2-6x+4 \\\\[-0.35em] ~\dotfill[/tex]
[tex]a(x^2-6x+4)=y\qquad \stackrel{\textit{we also know that}}{x =3 \qquad y = 10}\qquad \implies a[(3)^2-6(3)+4]=10 \\\\\\ a(9-18+4)=10\implies -5a=10\implies a = \cfrac{10}{-5}\implies \underline{a = -2} \\\\\\ -2(x^2-6x+4)=y\implies \boxed{-2x^2+12x-8=y}[/tex]
Check the picture below.
