The largest value of $x$ that satisfies $\sqrt{x+1}=x-2$ can be written as $\dfrac{a+\sqrt{b}}{c}$ where $c$ has no common factor with $a$ and $\sqrt{b}$ and $b$ is not divisible by the square of any integer greater than 1. What is the value of $a+b+c$?
We start by attempting to square both sides of the given equation to get rid of the radical.
$$
(\sqrt{x+1})^2 = (\sqrt{x+1})^2 \implies x+1 = x^2 - 4x + 4
$$
Simplifying, we have
$$
x^2 -5x +3 = 0
$$
By factoring, we find
\begin{align*} (x-3)(x-1)&= 0 \\ x &= 3, 1 \end{align*}
Let's check to find that $x=1$ doesn't work, but $x=\boxed{3}$ does work. Checking this solution in the original equation, we get $\sqrt{3+1}=3-2$ which simplifies to $\sqrt4=1$, so this root works as well.
To find the largest value of $x$ that satisfies the equation $\sqrt{x+1}=x-2$, we can start by squaring both sides of the equation:
$(\sqrt{x+1})^2=(x-2)^2$
$x+1=(x-2)^2$
Expanding the right side of the equation:
$x+1=x^2-4x+4$
Rearranging and simplifying the equation:
$x^2-5x+3=0$
We can solve this quadratic equation by factoring or by using the quadratic formula. Let's use the quadratic formula:
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
For our equation, $a=1$, $b=-5$, and $c=3$. Substituting these values into the quadratic formula:
$x=\dfrac{-(-5)\pm\sqrt{(-5)^2-4(1)(3)}}{2(1)}$
Simplifying the equation:
$x=\dfrac{5\pm\sqrt{25-12}}{2}$
$x=\dfrac{5\pm\sqrt{13}}{2}$
Since we are looking for the largest value of $x$, we take the positive square root:
$x=\dfrac{5+\sqrt{13}}{2}$
Therefore, $a=5$, $b=13$, and $c=2$. The value of $a+b+c$ is $5+13+2=\boxed{20}$.