135. A troop of monkeys was having fun; the square of one eighth of them was frolicking in the forest, the remaining twelve were shouting at the top of a hillock. Tell me: how many monkeys were there in total?
135. The equation will be $\left(\frac{1}{8} x\right)^{2}+12=x$, from which $x_{1}=48$; $x_{2}=16$

The equation, ready for solving, has the form: $x^{2}-64 x+768=0$, but Bhaskara writes $x^{2}-64 x=-768$ and adds $32^{2}$ to both sides:

$$
\begin{aligned}
& x^{2}-64 x+32^{2}=-768+1024 \\
& (x-32)^{2}=256 \\
& x-32=\sqrt{256}=16 \text{ and } x=48
\end{aligned}
$$

In this case, says Bhaskara, the negative units of the first part are such that the units of the second part are less than them, $10 \%$
and therefore the latter can be considered both positive and negative, and we obtain a double value of $x: 48$ and 16.