The quadratic $-6x^2+36x+216$ can be written in the form $a(x+b)^2+c$, where $a$, $b$, and $c$ are constants. What is $a+b+c$?
To write the quadratic in the given form, I need to complete the square. First, I factor out the coefficient of $x^2$, which is $-6$, from the first two terms: $-6x^2+36x+216=-6(x^2-6x)+216$. Next, I add and subtract the square of half of the coefficient of $x$, which is $(-6/2)^2=9$, inside the parentheses: $-6(x^2-6x)+216=-6(x^2-6x+9-9)+216$. Then, I group the first three terms inside the parentheses, which form a perfect square, and simplify the rest: $-6(x^2-6x+9-9)+216=-6((x-3)^2-9)+216$. Finally, I distribute the $-6$ and combine the constant terms: $-6((x-3)^2-9)+216=-6(x-3)^2+54+216$. Now I can see that the quadratic is in the form $a(x+b)^2+c$, where $a=-6$, $b=-3$, and $c=270$. To find $a+b+c$, I just add these values: $-6+(-3)+270=261$. 