Решите уравнения √2-x·√x+6=2(x+4)

[latex]\sqrt{2-x}\sqrt{x+6}=2(x+4)\\\\ x \leq 2\\ x \geq -6\\\\ \sqrt{-x^2-4x+12}=2(x+4)\\\\ -x^2-4x+12=2^2(x+4)^2\\\\ -x^2-4x+12=4(x^2+8x+16)\\\\ -x^2-4x+12=4x^2+32x+64\\\\ 5x^2+36x+52=0\\\\ D=36^2-4*5*52=16^2\\\\ x=\frac{-36+16}{10}=-2\\\\ x=\frac{-36-16}{10}=-\frac{26}{5}\\\\ [/latex]         Ответ [latex] x=-2[/latex]