x-intercepts of parabola

Question

y = -0.33x^2 + 0.84x + 6

x - intercepts of the parabola?

Answer 1

-.33x^2 + .84x + 6 = 0

x=.84/.66 + or - (1/.66)sqr(.84^2 +24(.33)) = .84/.66 + or - (1/.66)sqr(8.6256) = 84/66 + 2.937/.66= 3.777/.66 =5.723 or -3.177

x = about 5.72 or -3.18

those are the x intercepts, the values of x that make the quadratic = 0

the points (-3.18, 0) and (5.72, 0)

axis of symmetry is x=1.27 The x intercepts are equidistant to (1.27,0)