Yesterday there were $3t$ boys and $2t$ girls in the playground. Today, there are $3t-6$ boys and $2t-7$ girls. From $3t-6 = (2t-7)^2$ we get $4t^2-31t+55 = 0$, so $t_{1,2} = \frac{31\pm\sqrt{81}}{8}$. The only possible solution is $t = 5$, since $t$ is an integer. Yesterday there were $3t = 15$ boys and $2t = 10$ girls in the playground, or $25$ children altogether.