does this quadratic equation:

3x^2 - 2x - 9 = 0

have two solutions?? or none?

and if there are solutions will it be either 1.431 or -1.431 (correct 2 3 decimal places) or 4

also 2.097 remember there is a + or - sign in the equation. You have to try both.

To determine the number of solutions for a quadratic equation, you can use the discriminant (denoted as "D") which is part of the quadratic formula. The discriminant is calculated as follows:

D = b^2 - 4ac

Where:
- "a" is the coefficient of the x^2 term
- "b" is the coefficient of the x term
- "c" is the constant term

For the quadratic equation you provided: 3x^2 - 2x - 9 = 0
- "a" is 3
- "b" is -2
- "c" is -9

Now, let's calculate the discriminant:

D = (-2)^2 - 4 * 3 * (-9)
D = 4 + 108
D = 112

Since the discriminant is greater than zero (D > 0), the quadratic equation has two distinct real solutions.

Now, let's find those solutions:

x = (-b ± √D) / 2a

Substituting the values we have:

x = (-(-2) ± √112) / (2 * 3)
x = (2 ± √112) / 6
x = (2 ± √(16 * 7)) / 6
x = (2 ± 4√7) / 6
x = (1/3) ± (2/3)√7

To find the solutions rounded to three decimal places, we have:
- Solution 1: x = (1/3) + (2/3)√7 ≈ 1.431
- Solution 2: x = (1/3) - (2/3)√7 ≈ -1.431

Therefore, the solutions to the quadratic equation 3x^2 - 2x - 9 = 0, rounded to three decimal places, are approximately 1.431 (either positive or negative) and 4 is not a solution.