\( 2 r(t+3)+t^{2}=(t-5)^{2} \quad \) for \( t \)

To solve the equation \( 2 r(t+3)+t^{2}=(t-5)^{2} \), we first expand the right side. The right side becomes \( (t-5)^{2} = t^{2} - 10t + 25 \). Now, substituting this back into the equation gives us: \[ 2r(t+3) + t^{2} = t^{2} - 10t + 25. \] Next, we can subtract \( t^{2} \) from both sides, simplifying our equation to: \[ 2r(t+3) = -10t + 25. \] Now, divide both sides by \( 2 \): \[ r(t+3) = -5t + \frac{25}{2}. \] From here, you can solve for \( r \): \[ r = \frac{-5t + \frac{25}{2}}{t+3}. \] Now you've expressed \( r \) in terms of \( t \). Fun fact: you can explore how this relationship plays out for different values of \( t \) to see how \( r \) varies!