2230.11 – DON'T GUESS!

Playing it straight, multiply out:

$$\begin{align*}(y + 1) (2y - 3) &= 25 \\2 y^2 - 3y + 2y - 3 &= 25 \\2 y^2 - y - 28 &= 0 \\(y - 4) (2y + 7) &= 0\end{align*}$$

So $y$ is either $4$ or $\dfrac{-7}{2}$.

We might have hoped, looking at the original equation and noticing that $25 = 5 \times 5$, that perhaps $y + 1 = 5$ and $2y - 3 = 5$. This would give us $y = 4$ but misses $y = -7/2$.