To solve the quadratic equation (2x^2 - 6x + 3 = 0), you can use the quadratic formula:
[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
]
Here, (a = 2), (b = -6), and (c = 3).
First, calculate (b^2 - 4ac):
[
b^2 = (-6)^2 = 36
]
[
4ac = 4 \cdot 2 \cdot 3 = 24
]
[
b^2 - 4ac = 36 - 24 = 12
]
Now, substitute (a), (b), and (\sqrt{{b^2 - 4ac}}) into the quadratic formula:
[
x = \frac{{-(-6) \pm \sqrt{12}}}{2 \cdot 2}
]
[
x = \frac{{6 \pm \sqrt{12}}}{4}
]
Simplify (\sqrt{12}):
[
\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}
]
Substituting this back in:
[
x = \frac{{6 \pm 2\sqrt{3}}}{4}
]
[
x = \frac{6}{4} \pm \frac{2\sqrt{3}}{4}
]
[
x = \frac{3}{2} \pm \frac{\sqrt{3}}{2}
]
Thus, the solutions are:
[
x = \frac{3 + \sqrt{3}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{3}}{2}
]
These are the two roots of the quadratic equation (2x^2 - 6x + 3 = 0).
