Find the roots of the quadratic equation: $2{\mathrm{x}}^2-6\mathrm{x}+3=0$ 

Given a quadratic equation, $2{\mathrm{x}}^2-6\mathrm{x}+3=0$.............(i) 

On comparing eq (i) with ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$, we get: a = 2, b = - 6, c = 3 

Discriminant of a quadratic equation: $\mathrm{D}={\mathrm{b}}^2-4\mathrm{ac}$ 

Putting the value of a, b, c, we get:

$\begin{array}{r}\mathrm{D}=(-6{)}^2-4\times 2\times 3\\ \mathrm{D}=36-24\\ \mathrm{D}=12>0\end{array}$ 

Hence, the roots of the given quadratic equation are real and unequal.