SOLUTION: Find the x-intercepts for the parabola defined by the equation: y=-2x^2-4x+6.

Hi,         

Find the x-intercepts for the parabola y=-2x^2-4x+6

  -2x^2-4x+6 = 0

  -2(x^2+2x-3)=0 

 (x+3)(x-1)=0 Note:SUM of the inner product(3x) and the outer product(-x) = 2x

 (x+3)= 0

  x = -3

(x-1)=0

  x = 1

please note: Vertex is Pt(-1,8) 

as square completed for y=-2x^2-4x+6  gives y= -2(x+1)^2+ 8

the vertex form of a parabola,  where(h,k) is the vertex