2x^2-6x+3=0

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 \(2x^2 - 6x + 3 = 0\).