\[\sqrt{x^{2} + 3x} \leq 2\]
\[ОДЗ:x^{2} + 3x \geq 0\]
\[x(x + 3) \geq 0\]
\[x \leq - 3;\ \ x \geq 0.\]
\[x^{2} + 3x \leq 4\]
\[x^{2} + 3x - 4 \leq 0\]
\[x_{1} + x_{2} = - 3;x_{1} \cdot x_{2} = - 4\]
\[x_{1} = - 4;\ \ x_{2} = 1.\]
\[(x + 4)(x - 1) \leq 0\]
\[- 4 \leq x \leq 1.\]
\[Решение\ неравенства\ с\ учетом\ ОДЗ:\]
\[- 4 \leq x \leq - 3;\ \]
\[0 \leq x \leq 1.\]