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To find the zeros of this quadratic function, y= 3x^2 + 6x - 9, we must equal y to 0. So we have the quadratic equation:

3x^2+6x-9 = 0, where a = 3, b = 6, and c = -9

The quadratic formula:

x = [-b ± √(b^2 - 4ac)]/(2a) substitute what you know into this formula;

x = [-6 ± √(6^2 - 4 x 3 x -9)]/(2 x 3)

x = [-6 ± √(36 +108)]/6

x = (-6 ± √144)/6

x = (-6 ± 12)/6 Simplify: mulyiply by 1/6 both the numerator and the denominator;

x = -1 ± 2

x = -1 + 2 or x = -1 - 2

x = 1 or x = -3

So solutions are -3 and 1.

If you check the answers by plugging them into the equation, you will see that they work.

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Q: Solve y 3x2 6x-9 using the quadratic formula?
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