2. A weight with a mass of 200 grams stands on a table. It was flipped and placed on the table with a different side, the area of which is 15 sq. cm smaller. As a result, the pressure on the table increased by 1200 Pa. Find the area of the side on which the weight initially stood. Give your answer in sq. cm, rounding to two decimal places if necessary.

$\{25\}$
Solution. After converting to SI units, we get: $\frac{2}{S-1.5 \cdot 10^{-3}}-\frac{2}{S}=1200$.

Here $S-$ is the area of the original face.

From this, we get a quadratic equation: $4 \cdot 10^{5} S^{2}-600 S-1=0$.

After substituting the variable $y=200 S$, the equation becomes: $10 y^{2}-3 y-1=0$, the solution of which is easily found $y=1 / 2$. Thus, the area of the required face is $S=\frac{1}{400} \mathrm{~m}^{2}=25 \mathrm{~cm}^{2}$.

Answer: 25