For what values of x is \(2x^2+8x\le-6\)? Express your answer in interval notation.
+36915 2x^2+8x+6 <= 0
x^2 + 4x+3<=0
(x+1)(x+3)<=0
x = -1 and -3 for the equation to EQUAL zero
Now from - infinity to -3 it is greater than 0 and from -1 to + infinity it is greater than zero
so the only interval is from -3 to -1 that satisfies the equation
[-3,-1]
