y = 3x^2 - 3x - 6 {the x^2 (x squared) makes it a quadratic formula, and I'm assuming this is what you meant...}
This is derived from:
y = ax^2 + bx + c
So, by using the 'sum and product' rule:
a × c = 3 × (-6) = -18
b = -3
Now, we find the 'sum' and the 'product' of these two numbers, where b is the 'sum' and a × c is the 'product':
The two numbers are: -6 and 3
Proof:
-6 × 3 = -18 {product}
-6 + 3 = -3 {sum}
Now, since a > 1, we divide a from the results
-6/a = -6/3 = -2
3/a = 3/3 = 1
We then implement these numbers into our equation:
(x - 2) × (x + 1) = 0 {derived from 3x^2 - 3x - 6 = 0}
To find x, we make x the subject of 0:
x - 2 = 0
OR
x + 1 = 0
Therefore:
x = 2
OR
x = -1
So the x-intercepts of the quadratic formula (or solutions to equation 3x^2 - 3x -6 = 0, to put it into your words) are 2 and -1.
We can check this by substituting the values for x:
Let's start with x = 2:
y = 3(2)^2 - 3(2) - 6
= 3(4) - 6 - 6
= 12 - 6 - 6
= 0 {so when x = 2, y = 0, which is correct}
For when x = -1:
y = 3(-1)^2 - 3(-1) - 6
= 3(1) + 3 - 6
= 3 + 3 - 6
= 0 {so when x = -1, y = 0, which is correct}