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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If you mean 3x2+4x-2 = 0 then it can be solved by means of the quadratic equation formulla
-b +/- root b2 - 4ac / 2a(a = 3, b = -7, c = -3)7 +/- root -72 - 4 x 3 x -3 / 2 x 37 +/- root 49 + 36 / 67 +/- root 85 / 6
3x2-2x-2 = (3x-3.645751311)(x+0.5485837704) when factored with the help of the quadratic equation formula
The discriminant is -32.
6^2 -4(3*30) = -96 meaning that the given quadratic expression has no real roots