Step 1: Compare $2x^{2}+kx+3=0$ with $ax^{2}+bx+c=0$: $a=2,\; b=k,\; c=3$.
Step 2: For real and equal roots, the discriminant must be zero.
$$D=b^{2}-4ac=0$$
Step 3: Substitute the values.
$$k^{2}-4(2)(3)=0 \;\Rightarrow\; k^{2}-24=0$$
Step 4: Solve for $k$.
$$k^{2}=24 \;\Rightarrow\; k=\pm 2\sqrt{6}$$
Step 5: For equal roots, each root is $x=-\dfrac{b}{2a}=-\dfrac{k}{4}$.
Step 6: Case I — when $k=2\sqrt{6}$:
$$x=-\frac{2\sqrt{6}}{4}=-\frac{\sqrt{6}}{2}\quad(\text{repeated root})$$
Step 7: Case II — when $k=-2\sqrt{6}$:
$$x=-\frac{-2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}\quad(\text{repeated root})$$
Hence $k=\pm 2\sqrt{6}$, and the roots are $-\dfrac{\sqrt{6}}{2},\,-\dfrac{\sqrt{6}}{2}$ for $k=2\sqrt{6}$ and $\dfrac{\sqrt{6}}{2},\,\dfrac{\sqrt{6}}{2}$ for $k=-2\sqrt{6}$.