+128472 When the expression $-2x^2-20x-53$ is written in the form $a(x+d)^2+e$, where $a$, $d$, and $e$ are constants, then what is the sum $a+d+e$?
-2x^2 - 20x - 53 we need to complete the square on x...factor out the "-2"
-2 (x^2 + 10x + 53/2)
Take 1/2 of 10 = 5....square it = 25....add and subtract it inside the parentheses
-2 (x^2 +10x + 25 + 53/2 - 25) factor the first three terms and simplify the rest
-2 [ ( x + 5)^2 + 3/2) distribute the "-2" back across the terms inside the parentheses
-2(x + 5)^2 - 3
So a = -2, d = 5 and e = -3 and their sum is 0
