Решить уравнение x+((√12-x)/3))-12=0

$x+ \frac{ \sqrt{12-x} }{3} -12=0 \\ 12-x \geq 0~~~~~~x \leq 12 \\ 3*x+\sqrt{12-x}-12*3=0\\ 3x+\sqrt{12-x}-36=0 \\ \sqrt{12-x}=36-3x \\ (\sqrt{12-x})^2=(36-3x )^2 \\ 12-x=1296-216x+9x^2 \\ 9x^2-216x+x+1296-12=0 \\ 9x^2-215x+1284=0 \\ D=(-215)^2-4*9*1284=46225-46224=1 \\ x_1= \frac{-(-215)-\sqrt{1}}{2*9} = \frac{215-1}{18} = \frac{214}{18}= \frac{107}{9}=11 \frac{8}{9} \\ x_2= \frac{-(-215)+\sqrt{1}}{2*9} = \frac{215+1}{18} = \frac{216}{18} =12$

Проверка:
$11 \frac{8}{9} + \frac{ \sqrt{12-11 \frac{8}{9} } }{3} -12=11 \frac{8}{9}-11 \frac{9}{9} +\frac{ \sqrt{11 \frac{9}{9} -11 \frac{8}{9} } }{3}=- \frac{1}{9} + \frac{ \sqrt{ \frac{1}{9}} }{3} = \\ =- \frac{1}{9} + \frac{ \frac{1}{3} }{3} = - \frac{1}{9} +- \frac{1}{9} =0 \\ \\ 12+ \frac{ \sqrt{12-12} }{3} -12= 12+ \frac{ \sqrt{0} }{3} -12=12+0-12=0$