Find the slope of the tangent to the curve `x=t^2+3t-8` , `y=2t^2-2t-5` at `t=2` .

Explanation

To find the slope of the tangent to the curve defined by the parametric equations \( x = t^2 + 3t - 8 \) and \( y = 2t^2 - 2t - 5 \) at \( t = 2 \), we will follow these steps:

### Step 1: Differentiate \( x \) and \( y \) with respect to \( t \)

First, we need to find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

1. For \( x = t^2 + 3t - 8 \):
\[
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