Количество целых решений неравенства: х^2+14х+49<5 |х+7|

${x}^{2} + 14x + 49 < 5 \times |x + 7| \\ 1)x + 7 \geqslant 0 \: \: \: x \geqslant -7 \\ {x}^{2} + 14x + 49 < 5 \times (x + 7) \\ {x}^{2} + 14x + 49 < 5x + 35 \\ {x}^{2} + 9x + 14 < 0 \\ {x }^{2} + 9x + 14 = 0 \\ d = 81 - 4 \times 14 = 81 - 54 = 25 = {5}^{2} \\ x = \frac{ - 9±5}{2} = - 2 \: \: - 7 \\ + + + + ( - 7) - - - ( - 2) + + \\ x\in( - 7; - 2) \\ 2)x + 7 < 0 \: \: \: x < - 7 \\ {x }^{2} + 14x + 49 < 5 \times - (x + 7) \\ {x}^{2} + 14x + 49 < - 5x - 35 \\ {x}^{2} + 19x + 84 < 0 \\ {x}^{2} + 19x + 84 = 0 \\ d \: = 361 - 4 \times 84 = 361 - 336 = 25 = {5}^{2} \\ x = \frac{ - 19 ±5}{2} = - 12; - 7 \\ + + + + ( - 12) - - - ( - 7) + + + \\ x\in( - 12;- 7)$
Ответ :
$x\in( - 12; - 7)\cup ( - 7; - 2)$
Количество целых решений : 8