What is the axis of h(x) = 5x^2+40x+64

H (x) = 5x^2+40x+64
y = (5x^2+40x+60) + 4
y = 5(x^2+8x+12) + 4
y = 5(x+2)(x+6) + 4
y-4 = 5(x+2)(x+6)
This means that the solutions, or where the parabola crosses the x-axis, are at x+2=0 and x+6=0, or x=-2, -6
So the midpoint of -2 and -6 is -4, the axis of symmetry is x=-4