jmc

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 $$a{x}^2+bx+c=-(x-2{)}^2+9.$$ In order to compute $100a+10b+c$ we can substitute $x=10$ and that gives $$100a+10b+c=-(10-2{)}^2+9=\boxed{-55}.$$