Meeting Point

We note that the $LHS = \left(\dfrac 9{10}\right)^x > 0$ for all $x$ . Now consider the $RHS$ as follows.

$\begin{aligned} RHS & = - 3 + x - x^2 \\ & = - \left(x^2 - x + 3\right) \\ & = - \left(x - \frac 12 \right)^2 - \frac {11}4 & \small \color{#3D99F6} \text{Since }\left(x^2 - \frac 12 \right)^2 \ge 0 \\ \implies RHS & \le - \frac {11}4 < 0 \end{aligned}$

This implies that $LHS \ne RHS$ , therefore, there is $\boxed{0}$ solution to the equation.

$(0.9)^x$ = $-3 + x - x^2$

Multiply both sides by $-1$

$-(0.9)^x$ $= 3 - x + x^2$

$-(0.9)^x$ $=(x-0.5)^2 + 2.75$

Notice that the right-hand side is always positive and the left-hand side is always negative (because a positive number raised to the power of any real $x$ is always positive). So there are no real solutions.