You know the values of the two roots, and we know that one form of the quadratic for this will be given by
$\displaystyle (x - r_1)(x - r_2)$
Thus
$\displaystyle (x - (3 + \sqrt{2} ))(x - (3 - \sqrt{2}))$
The simplest way to expand this (in my opinion) is to write this as
$\displaystyle ((x - 3) - \sqrt{2} ) ((x - 3) + \sqrt{2})$
which is in the form of $\displaystyle (a - b)(a + b) = a^2 - b^2$,
$\displaystyle ((x - 3) - \sqrt{2} ) ((x - 3) + \sqrt{2}) = (x - 3)^2 - (\sqrt{2})^2$
$\displaystyle = x^2 - 6x + 9 - 2 = x^2 - 6x + 7$
as you can check.
-Dan
Substitute the given values into the general equation, and you will end up with a system of linear equations in $\displaystyle a,\;b,\text{ and }c$:
$\displaystyle \begin{array}{rcl}
(x,\;y) &\rightarrow& y = ax^2 + bx + c\\
(0,\;8) &\rightarrow& 8 = a(0)^2 + b(0) + c\\
(8,\;0) &\rightarrow& 0 = a(8)^2 + b(8) + c\\
(-2,\;5) &\rightarrow& 5 = a(-2)^2 + b(-2) + c
\end{array}$
and you get the following system:
$\displaystyle \left\{\begin{array}{rcrcrcr}
64a & + & 8b & + & c & = & 0\\
4a & - & 2b & + & c & = & 5\\
&&&& c & = & 8
\end{array}\right.$
Now solve for the needed coefficients.