81. The formula is derived exactly as in ordinary algebra.

We use the identity

$$a x^{2}+b x+c=a\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a}$$

By this identity, the equation

$$a x^{2}+b x+c=0$$

is equivalent to the equality

$$a\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a}=0$$

or

$$\left(x+\frac{b}{2 a}\right)^{2}=\frac{b^{2}-4 a c}{4 a^{2}}$$

From the last equality, it is clear that if equation (1) has solutions, then the square root of $b^{2} - 4ac$ must be extractable. Equality (2) can be rewritten in this case as

$$x+\frac{b}{2 a}=\frac{\pm \sqrt{b^{2}-4 a c}}{2 a}$$

from which

$$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$

Thus, equation (1) has no solutions if $\sqrt{b^{2}-4 a c}$ cannot be extracted, and has two solutions, calculable by formula (3), if $\sqrt{b^{2}-4 a c}$ can be extracted. These solutions are distinct if $b^{2}-4 a c \neq 0$, and coincide if $b^{2}-4 a c=0$. In the case of 2-arithmetic, formula (3) loses its meaning, as it contains division by two.