Вопрос в картинках...

Решите задачу:

$(x^2+14x+24)(x^2+11x+24)=4x^2 \\ x^2(x+14+\frac{24}{x})(x+11+\frac{24}{x})=4x^2 |:x^2 \\ (x+\frac{24}{x}+14)(x+\frac{24}{x}+11)=4 \\ x+\frac{24}{x} = t \\ (t+14)(t+11)=4 \\ t^2+11t+14t+154-4=0 \\ t^2+25t+150=0 \\ D=25^2-600 = 25 \\ t_{1,2} = \frac{-25\pm5}{2} \\ t_1 = \frac{-25-5}{2} = -15 \\ t_2 = \frac{-25+5}{2} = -10 \\ x+\frac{24}{x} = -10 | * x \\ x^2+10x+24=0 \\ D = 100-24*4 = 4 \\ x_{1,2} = \frac{-10\pm2}{2} \\ x_1 = -6; x_2 = -4 \\ x+\frac{24}{x} = -15 \\ x^2+15x+24=0 \\ D = 225-24*4 = 129 \\$
$x_{3,4} = \frac{-15\pm\sqrt{129}}{2} \\$

$Ans: x \in \{ -6, -4, \frac{-15-\sqrt{129}}{2}, \frac{-15+\sqrt{129}}{2}\}$