Why is the LaTeX document not compiling?

Question

TeX

I'm practicing in LaTeX and ran into the following problem:

! Missing $ inserted.
<inserted text> 
                $
l.28 		\begin{flushleft}

the code

\documentclass[a4paper,12pt]{report}
\usepackage[utf8]{inputenc}
\usepackage[T2A]{fontenc}
\usepackage{textcomp}
\usepackage[english,russian]{babel}
\usepackage{amsmath, amssymb}


\usepackage{geometry} % Меняем поля страницы
\geometry{left=2cm}% левое поле
\geometry{right=1.5cm}% правое поле
\geometry{top=1cm}% верхнее поле
\geometry{bottom=2cm}% нижнее поле

\begin{document}


  30.4 в \\
  $\sqrt{\frac{5x-1}{x+3}} =  2$ \\
  $\frac{5x-1}{x+3} = 4, x\neq -3$ \\
  $5x-1 = 4x+12$ \\
  $x = 11$ \\  
  Ответ: 11.\\ \\
  30.6 в \\
  
  $\sqrt{3x+4} = \sqrt{5x+2} $
  \begin{equation*}
    \begin{flushleft}
  \begin{cases}
      3x+4 \ge 0 \\
      5x+2 \ge 0
  \end{cases}
  \begin{cases}
    x\ge -1.25 \\ 
    x \ge  -0.4
  \end{cases}
  \rightarrow D(x) \in [-0.4; +\infty) \\
\end{flushleft}
\end{equation*}

  $3x+4=5x+2$ \\
  $2 = 2x$ \\
   $x = 1$ \\
   $x \in D(x)$ \\
   Ответ: 1. \\ \\
   30.11 в \\

   $\sqrt{15+3x} = 1-x,    x\le 1 \Rightarrow D(x) \in (-\infty; 1] $\\
  $15+3x = x^2-2x+1$ \\
  $x^2 - 5x - 14 = 0$ \\
  $D = 25+4 \cdot 14 = 25 + 56 = 81$ \\
  $x_1 = \frac{5+9}{2} = 7$ \\
  $x_2 = \frac{5-9}{2} = -2$ \\
  $x_1 \not\in D(x)$ \\
  Ответ: -2. \\ \\
  30.18 в \\
  $\sqrt{x^2+x+1} = x+2, x\ge -2 \rightarrow D(x) \in [-2; +\infty) $ \\
  $x^2+x+1 = x^2 + 4x + 4$ \\
  $3x = -3$ \\
   $x = -1$ \\
   Ответ: -1. \\ \\
   30.21 в \\
   $\sqrt{3x+1} + \sqrt{x-4} = 2\sqrt{x} $ 
   \begin{equation*}
     \begin{flushleft}
   	\begin{cases}
   		3x+1 \ge  0 \\
      x-4 \ge  0 \\
      x \ge  0 \\
   	\end{cases}
    \begin{cases}
      x \ge  -\frac{1}{3} \\
      x \ge  4 \\
      x \ge  0\\
    \end{cases}
    \Rightarrow D(x) \in [4; +\infty) \\
  \end{flushleft}
   \end{equation*}
   $3x+1+x-4 +2\sqrt{(3x+1)(x-4)} = 4x$ \\
   $4x-3+2\sqrt{3x^2-11x -4 } = 4x $ \\
   $2\sqrt{3x^2-11x-4}=3 $ 
   $12x^2-44x-16=9$ \\
   $12x^2-44x-25 = 0$ \\
   $D_1 = 22^2 + 12 \cdot 25 = 484 +300 = 784$ \\
   $x_1 = \frac{22+28}{12} = \frac{25}{6} = 4\frac{1}{6}$ \\
   $x_2 = \frac{22-28}{12} = -\frac{1}{2}$ \\
   $x_2 \not\in D(x)$ \\
   Ответ: $4\frac{1}{6}$\\
\end{document}

I tried to put $$ in systems of equations, but it didn't help.
Why such an error? And how can the solution be indented so that it can be visually distinguished from the task number?

Answer 1

\begin{flushleft}
this is what needs to be removed.