Vertex form of a parabola
y = a (x - h)^2 + k
where (h, k) is the vertex
Substituting the values of h and k.
we get,
y = a(x + 4)^2 + 2
substituting in the point (0, -30) for x and y
-30 = a (0 + 4)^2 + 2
solve for a,
-30 = 16 a + 2
-32 = 16 a
-2 = a
y = -2(x + 4)^2 + 2
Put y = 0
-2 x^2 - 16 x - 30 = 0
-2(x^2 + 8 x + 15) = 0
x^2 + 8 x + 15 = 0
(x + 3)(x + 5) = 0
x = -3
x = -5
