Since the points $ (-1,0) $ and $ (5,0) $ have the same $ y $-value, the axis of symmetry of the parabola must be between these 2 points. The $ x $-value halfway between $ -1 $ and $ 5 $ is $ x=2 $. Therefore the vertex of the parabola is equal to $ (2,k) $ for some $ k $ and the parabola may also be written as \[a(x-2)^2+k.\] Now we substitute. The point $ (5,0) $ gives \[0=a(5-2)^2+k,\] or \[9a+k=0.\] The point $ (0,5) $ gives \[5=a(0-2)^2+k\] or \[4a+k=5.\] Subtracting the second equation from the first gives \[(9a+k)-(4a+k)=0-5\] so $ 5a=-5 $, giving $ a=-1 $. Since $ a=-1 $ and $ 9a+k=0 $ we know that $ k=9 $ and our parabola is \[ax^2+bx+c=-(x-2)^2+9.\] In order to compute $ 100a+10b+c $ we can substitute $ x=10 $ and that gives \[100a+10b+c=-(10-2)^2+9=\boxed{-55}.\]