#### Answer

The solutions are:
$\\x_1\approx 2.4$
$\\x_2 \approx 0.6$

#### Work Step by Step

RECALL:
A quadratic equation of the form $ax^2+bx+c=0$ can be solved using the quadratic formula
$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
The quadratic equation $2x^2-6x+3=0$has:
$a=2
\\b=-6
\\c=3$
Substitute these values into the quadratic formula to obtain:
$x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(2)(3)}}{2(2)}
\\x=\dfrac{6 \pm \sqrt{36-24}}{4}
\\x=\dfrac{6\pm \sqrt{12}}{4}
\\x =\dfrac{6}{4} \pm \dfrac{\sqrt{12}}{4}
\\x=\dfrac{3}{2} \pm \dfrac{\sqrt{12}}{4}$
Thus,
$\\x_1=\dfrac{3}{2}+\dfrac{\sqrt{12}}{4} \approx 2.4$
$\\x_2=\dfrac{3}{2}-\dfrac{\sqrt{12}}{4} \approx 0.6$