To solve the quadratic equation ( -3X^2 + X + 2 = 0 ), we can rearrange it into a standard form:
[
3X^2 - X - 2 = 0
]
Now we can use the quadratic formula, which is given by:
[
X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
]
Here, ( a = 3 ), ( b = -1 ), and ( c = -2 ).
First, we calculate the discriminant ( b^2 - 4ac ):
[
b^2 = (-1)^2 = 1
]
[
4ac = 4 \cdot 3 \cdot (-2) = -24
]
[
b^2 - 4ac = 1 - (-24) = 1 + 24 = 25
]
Now we can plug the values into the quadratic formula:
[
X = \frac{-(-1) \pm \sqrt{25}}{2 \cdot 3}
]
[
X = \frac{1 \pm 5}{6}
]
This gives us two potential solutions:
When we take the positive root:
[
X = \frac{1 + 5}{6} = \frac{6}{6} = 1
]
When we take the negative root:
[
X = \frac{1 - 5}{6} = \frac{-4}{6} = -\frac{2}{3}
]
Thus, the solutions to the equation ( -3X^2 + X + 2 = 0 ) are:
[
X = 1 \quad \text{and} \quad X = -\frac{2}{3}
]
