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

For the point $(2,-1)$ on the curve

$x=t^2+3t-8,y=2t^2-2t-5,$ we have

$\begin{array}{r}{t}^2+3t-8=2\text{ and }2t^2-2t-5=-1\\ \Rightarrow t^2+3t-10=0\text{ and }2t^2-2t-4=0\\ \Rightarrow (t+5)(t-2)=0\text{ and }(t-2)(t+1)=0\\ \Rightarrow t=2\end{array}$

Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{dx}=\frac{4t-2}{2t+3}\Rightarrow {\left(\frac{dy}{dx}\right)}_{t=2}=\frac{4\times 2-2}{2\times 2+3}=\frac{6}{7}$