1. Determine the value of the parameter $k$ so that for the solutions $x_{1}$ and $x_{2}$ of the equation

$$
2 x^{2}+3 x+3 k+1=0
$$

the following holds

$$
2 x_{1}-x_{2}=3
$$
1. From Vieta's formulas, we have $x_{1}+x_{2}=-\frac{3}{2}$, so we can write $3=$ $2 x_{1}-x_{2}=2\left(x_{1}+x_{2}\right)-3 x_{2}=-3-3 x_{2}$, and we calculate $x_{2}=-2$. Substituting $x=-2$ into the given equation, we have $2(-2)^{2}+3(-2)+3 k+1=0$, i.e., $3+3 k=0$, from which we obtain $k=-1$.