I tried to solve for the final velocities i can't see what i missed. Let the 🔴 $puck_{red}$ have $m_1$=15g be stationary and the 🔵 $puck_{blue}$ have mass $m_2$=12 g moving at $v_{2i}=2.5 m/s$
$m_1v_{1i}-m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$
$$ v_{1f} =\frac{m_1v_{1i}}{m_1}-\frac{m_2v_{2i}}{m_2}-\frac{m_2v_{2f}}{m_1} \\ =-\frac{m_2v_{2i}}{m_1}-\frac{m_2v_{2f}}{m_1} $$
*let $v_{1i}=0$ because the 🔴 $puck_{red}$ is stationary
$$ v_{2f} =-\frac{m_2v_{2i}}{m_2}-\frac{m_1v_{1f}}{m_2} \\ =-v_{2i}-\frac{m_1v_{1f}}{m_2} $$
$\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2$
$m_2v_{2i}^2=m_1v_{1f}^2+m_2v_{2f}^2$
$v_{1f}^2 = \frac{m_2v_{2i}^2-m_2v_{2f}^2}{m_1}$
$v_{2f}^2 = \frac{m_1v_{1f}^2-m_2v_{2i}^2}{m_2} = \frac{m_1v_{1f}^2}{m_2}-v_{2i}^2$
*we also know that by Newton's 2nd law that the red puck will apply a negative force to the blue puck upon collision. In the initial conditions we will see $v_{2i}$ moving in the negative direction.
🔴 $\leftarrow$ 🔵 $\leftarrow$ (-x)
the left is found by just solving for the final velocity and dividing the other side by the mass associated with the following velocity.
We will simplify the equation solving for $v_{1f}, v_{2f}$
*note that $v_{1i}=0$ here as well.
I can't really seem to eliminate at this point because everything appear to just cancel out in the case where I get
$$ v_{1f}=v_{1f} \\v_{2f}=v_{2f} $$