To solve the quadratic equation (3x^2 + 2x + 5 = 0), we can use the quadratic formula:
[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}
]
where (a = 3), (b = 2), and (c = 5).
First, we calculate the discriminant (b^2 - 4ac):
[
b^2 - 4ac = 2^2 - 4 \cdot 3 \cdot 5 = 4 - 60 = -56
]
Since the discriminant is negative, this means the equation has no real solutions (the solutions will be complex numbers).
Now, we can use the quadratic formula to find the complex solutions:
[
x = \frac{{-2 \pm \sqrt{{-56}}}}{{2 \cdot 3}}
]
We know that (\sqrt{-56} = \sqrt{56}i = \sqrt{4 \cdot 14}i = 2\sqrt{14}i).
Now we can substitute back into the formula:
[
x = \frac{{-2 \pm 2\sqrt{14}i}}{{6}} = \frac{{-1 \pm \sqrt{14}i}}{{3}}
]
Thus, the solutions to the equation (3x^2 + 2x + 5 = 0) are:
[
x = \frac{{-1 + \sqrt{14}i}}{3} \quad \text{and} \quad x = \frac{{-1 - \sqrt{14}i}}{3}
]
