1. Let $\{x\}$ and $[x]$ denote the fractional and integer parts of the number $x$. The integer part of the number $x$ is the greatest integer not exceeding $x$, and $\{x\}=x-[x]$. Find $x$ for which $4 x^{2}-5[x]+8\{x\}=19$.
Solution.

$$
\begin{aligned}
& x=[x]+\{x\} \rightarrow 4([x]+\{x\})^{2}-5[x]+8\{x\}=19 \rightarrow \\
& 4\{x\}^{2}+8([x]+1)\{x\}+4[x]^{2}-5[x]-19=0
\end{aligned}
$$

If $x$ is the desired value, then the quadratic trinomial $f(t)=4 t^{2}+8([x]+1) t+4[x]^{2}-5[x]-19$ has a root $t=\{x\}$ on the interval $[0 ; 1)$.

Case 1. One root on $[0 ; 1)$.

$$
\left\{\begin{array} { c } 
{ f ( 0 ) = 4 [ x ] ^ { 2 } - 5 [ x ] - 1 9 } \\
{ f ( 1 ) = 4 [ x ] ^ { 2 } + 3 [ x ] - 7 \neq 0 } \\
{ f ( 0 ) \cdot f ( 1 ) \leq 0 }
\end{array} \rightarrow \left\{\begin{array}{c}
\left([x]-t_{1}\right)\left([x]-t_{2}\right)([x]-1)(4[x]+7) \leq 0 \\
{[x] \neq 1, t_{1}=\frac{-5-\sqrt{329}}{8} \approx-1.6, t_{2}=\frac{-5+\sqrt{329}}{8} \approx 2.9}
\end{array}\right.\right.
$$

Solving the inequality using the interval method, we get $[x] \in\left(-\frac{7}{4} ; t_{1}\right] \cup\left(1 ; t_{2}\right]$. Considering the integer nature of $[x]$, we get the only possible value of the integer part of the number $x$: $[x]=2$. Substituting $[x]=2$ into ( ${ }^{*}$ ) to determine $\{x\}:$

$$
4 t^{2}+24 t-13=0 \rightarrow\{x\}=\frac{1}{2},\{x\} \neq-\frac{13}{2} \rightarrow x=\frac{5}{2}
$$

Case 2. Two roots on $[0 ; 1)$. It is necessary that the abscissa of the vertex of the parabola $f(t)$ belongs to $(0,1)$, i.e., $t_{*}=-[x]-1 \in(0,1) \rightarrow[x] \in(-2 ;-1)$. There are no integers in the specified interval, so Case 2 does not occur.

Answer: $x=\frac{5}{2}$.