Let $f(x)=-3+x-x^{2}$
Then, $f(x) < 0$ for all $x$ because coefficient of $\mathrm{x}^{2} < 0$ and disc $ < 0 .$ Thus, LHS of the given equation is always positive whereas the RHS is always less than zero.
Hence, the given equation has no solution.
Alternate Solution :
Given, equation is
$$
\frac{9}{10}=-3+x-x^{2}
$$

Let $y=\frac{9}{10}$, therefore
$$
\begin{array}{l}
y=-3+x-x^{2} \\
y=-\left[x^{2}-x+\frac{1}{4}\right]-3+\frac{1}{4} \\
\Rightarrow y+\frac{11}{4}=-\left(x-\frac{1}{2}\right)^{2}
\end{array}
$$
It is clear from the graph that two curves do not intersect. Hence, no solution exists.