The slope of the tangent to the curve $x=t^2+3 t-8, y=2 t^2-2 t-5$ at the point (2, -1), is

$\frac{6}{7}$

For the point (2, -1) on the curve $x=t^2+3 t-8, y=2 t^2-2 t-5$, we have

$t^2+3 t-8=2$ and $ 2 t^2-2 t-5=-1$

$\Rightarrow t^2+3 t-10=0$ and $ 2 t^2-2 t-4=0$

$\Rightarrow (t+5)(t-2)=0$ and $(t-2)(t+1)=0$

$\Rightarrow t=2$

Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{4 t-2}{2 t+3} \Rightarrow\left(\frac{d y}{d x}\right)_{t=2}=\frac{4 \times 2-2}{2 \times 2+3}=\frac{6}{7}$