A parabola with equation $y=ax^2+bx+c$ has a vertical line of symmetry at $x=1$ and goes through the two points $(-1,3)$ and $(2,-2)$. The quadratic $ax^2 + bx +c$  has two real roots. The greater root is $\sqrt{n}+1$. What is $n$?
Rewrite the equation of the parabola as $y=a(x-h)^2+k$, where $a$, $h$, and $k$ are constants and $(h,k)$ are the coordinates of the vertex. If the parabola has a vertical line of symmetry at $x=1$, then the $x$-coordinate of the vertex is $x=1$, so $h=1$. The equation of the parabola becomes $y=a(x-1)^2+k$. Plugging in the two given points into this equation, we have the two equations \begin{align*}
3&=a(-1-1)^2+k \Rightarrow 3=4a+k\\
-2&=a(2-1)^2+k \Rightarrow -2=a+k
\end{align*} Subtracting the second equation from the first yields $5=3a$, so $a=5/3$. Plugging this value into the second equation to solve for $k$, we find that $k=-11/3$. So the equation of the parabola is $y=\frac{5}{3}(x-1)^2-\frac{11}{3}$. To find the zeros of the parabola, we set $y=0$ and solve for $x$: \begin{align*}
0&=\frac{5}{3}(x-1)^2-\frac{11}{3}\\
\frac{11}{3}&=\frac{5}{3}(x-1)^2 &\\
\frac{11}{5}&=(x-1)^2\\
x &= \pm\sqrt{\frac{11}{5}}+1
\end{align*}

The greater zero is at $x=\sqrt{\frac{11}{5}}+1$, so $n=\boxed{2.2}$. The graph of the parabola is below:

[asy]
Label f;

f.p=fontsize(4);

xaxis(-1,3,Ticks(f, 1.0));

yaxis(-4,3,Ticks(f, 1.0));

real f(real x)

{

return 5/3*(x-1)^2-11/3;

}

draw(graph(f,-1,3));
[/asy]