so, the equation has a leading term with a negative coefficient, and is a quadratic, meaning is a parabola opening downwards, so it goes up and up and up and then makes a U-turn then down down, so it has a maximum point, that is, at the vertex point
so.. where the dickens is the vertex at?
we're asked to make the vertex with a y-coordinate of 104, so the maximum value is 104
well
[tex]\bf y = {{ a}}x^2{{ +b}}x{{ +c}}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)\\\\
-----------------------------\\\\
\textit{so we want }\implies {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}=104
\\\\
\textit{let's take a peek at your equation}\quad
\begin{array}{ccccllll}
f(x)=&-x^2&+bx&-17\\
f(x)=&-1x^2&+bx&-17\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\\\\
thus
\\\\
(-17)=\cfrac{b^2}{4(-1)}=104[/tex]
solve for "b"
