Let $f(x)=-3+x-x^{2}$
Then, $f(x)<0$ for all $x$ because coefficient of $x^{2}<0$ 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 Given equation is
$\frac{9}{10}=-3+x-x^{2}$
Let $y=\frac{9}{10}$, therefore
$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} $
It is clear from the graph that two curves do not intersect. Hence, no solution exist.
