If $\mathrm{x}=\left(-\frac{1}{2}\right)$, is a solution of the quadratic equation $3{\mathrm{x}}^2+2\mathrm{kx}-3=0$, find the value of k.

We need to find the value of k, if $\mathrm{x}=\left(-\frac{1}{2}\right)$, is a solution of the quadratic equation $3{\mathrm{x}}^2+2\mathrm{kx}-3=0$.

It is known that if we substitute the solution in place of variable in an equation, then it results to 0. 

Therefore, if f(x) represents the quadratic equation $3{\mathrm{x}}^2+2\mathrm{kx}-3=0$, then and $\mathrm{\alpha }$ is the solution of f(x), then f($\mathrm{\alpha }$)=0.

On substituting $\mathrm{x}=\left(-\frac{1}{2}\right)$ in $3{\mathrm{x}}^2+2\mathrm{kx}-3=0$, we get

$\begin{array}{l}\Rightarrow 3{\left(-\frac{1}{2}\right)}^2+2\mathrm{k}\left(-\frac{1}{2}\right)-3=0\\ \Rightarrow 3\left(\frac{1}{4}\right)+\mathrm{k}(-1)-3=0\\ \Rightarrow \frac{3}{4}-\mathrm{k}-3=0\\ \Rightarrow \mathrm{k}=\frac{3}{4}-3\\ \Rightarrow \mathrm{k}=-\frac{9}{4}\end{array}$

Hence, the value of k is -$\frac{9}{4}$.