2x^{2} – 6x + 3 = 0

Comparing the equation with ax^{2} + bx + c = 0, we get

a = 2, b = -6, c = 3

As we know, Discriminant = b^{2} – 4ac

= (-6)^{2} – 4 (2) (3)

= 36 – 24 = 12

As b^{2} – 4ac > 0,

Therefore, there are distinct real roots exist for this equation, 2x^{2} – 6x + 3 = 0.

\( x= \dfrac{-b±\sqrt{b^2-4ac}}{2a}\)

= \(x= \dfrac{(-6)±\sqrt{(-6)^2-4(2)(3)}}{2(2)} \)

= \(x= \dfrac{6±\sqrt{12}}{4}=\dfrac{6±2\sqrt{3}}{4} \)

= \(x= \dfrac{3±\sqrt{3}}{2} \)

Therefore the roots for the given equation are \( \dfrac{3+\sqrt{3}}{2} \;and\; \dfrac{3-\sqrt{3}}{2} \)

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