jmc

The square which agrees with $x^2-x-1$ except for the constant term is ${\left(x-\frac{1}{2}\right)}^2$, which is equal to $x^2-x+\frac{1}{4}$, and thus to $(x^2-x-1)+\frac{5}{4}$.

Therefore, by adding $\frac{5}{4}$ to each side, Sviatoslav rewrote the equation $x^2-x-1=0$ as $${\left(x-\frac{1}{2}\right)}^2=\frac{5}{4}.$$We have $a=-\frac{1}{2}$ and $b=\boxed{\frac{5}{4}}$.