(x^2+4x/7x-2)-(12-42x/x^2+4x)=7 Решите уравнение с заменой

Решите задачу:

$\frac{x^2+4x}{7x-2}-\frac{12-42x}{x^2+4x}=7\\\frac{x^2+4x}{7x-2}+\frac{6(7x-2)}{x^2+4x}=7\\\frac{x^2+4x}{7x-2}=y\\y+\frac{6}{y}=7|*y\\y^2-7y+6=0\\D:49-24=25\\t=\frac{7\pm 5}{2}\\\\t_1=6\\\frac{x^2+4x}{7x-2}=6\\x^2+4x=42x+12\\x^2-38x+12=0\\D:1444-48=1396\\x_1,_2=\frac{38\pm2\sqrt{349}}{2}=19\pm \sqrt{349}; \\\\t_2=1\\\frac{x^2+4x}{7x-2}=1\\x^2+4x=7x-2\\x^2+4x-7x+2=0\\x^2-3x+2=0\\D: 9-8=1\\x_1, _2=\frac{3\pm 1}{2}\\x_1=2;\\x_2=1$