Explicație pas cu pas:
[tex]{(2x - 3)}^{2} + {(3x + 2)}^{2} \geqslant {(x + 5)}^{2} \\ 4 {x}^{2} - 12x + 9 + 9 {x}^{2} + 12x + 4 \geqslant {x}^{2} + 10x + 25 \\ 12 {x}^{2} - 10x - 12 \geqslant 0 \iff 6 {x}^{2} - 5x - 6 \geqslant 0 \\ (3x + 2)(2x - 3) \geqslant 0[/tex]
[tex]3x + 2 = 0 \implies x = - \frac{2}{3} \\ 2x - 3 = 0 \implies x = \frac{3}{2} [/tex]
[tex]\implies \bf x \in \Big( - \infty ; - \frac{2}{3} \Big] \cup \Big[ \frac{3}{2} ; + \infty\Big) \\ [/tex]
