Solve the following equation:

$$
3 \cdot 4^{x}+(3 x-10) 2^{x}+3-x=0
$$
Solution. Let's introduce the new variable $2^{x}=a>0$. After substitution and rearrangement, we arrive at the following quadratic equation in $a$:

$$
3 a^{2}+(3 x-10) a+3-x=0
$$

From this,

$$
a_{1,2}=\frac{-(3 x-10) \pm \sqrt{(3 x-10)^{2}-12(3-x)}}{6}
$$

Performing the indicated operations under the square root, we get $(3 x-8)^{2}$. Thus,

$$
\begin{aligned}
& a_{1}=\frac{10-3 x+3 x-8}{6}=\frac{1}{3} \quad \text { and } \quad 2^{x}=\frac{1}{3} \text {, so } \quad x_{1}=\log _{2} \frac{1}{3} \approx-1.585 \\
& a_{2}=\frac{10-3 x-3 x+8}{6}=3-x, \quad \text { i.e., } \quad 2^{x}=3-x
\end{aligned}
$$

Let's plot these two functions in the coordinate system. The function $2^{x}$ is strictly increasing, while $3-x$ is strictly decreasing, so they can have at most one common point. By substitution, we can verify that this point is at $x=1$. Therefore, the solutions to the equation are: $x_{1}=\log _{2} \frac{1}{3}$ and $x_{2}=1$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d6e4dbe19e9c32709908g-1.jpg?height=510&width=501&top_left_y=935&top_left_x=801)