Notice that the factored form of the part under the radical is:
(x-3)(x-3) which is equal to:
(x-3)^2
√((x-3)^2)+3|(x-3)(2x+1)|=0
√((x-3)^2)=-3|(x-3)(2x+1)| now squaring both sides
(x-3)^2=9(x-3)^2*(2x+1)^2 dividing both sides by (x-3)^2
1=9(2x+1)^2
1=9(4x^2+4x+1)
1=36x^2+36x+9
36x^2+36x+8=0
36x^2+36x=-8
x^2+x=-2/9
x^2+x+1/4=1/4-2/9
(x+1/2)^2=(9-8)/36
(x+1/2)^2=1/36 taking the square root of both sides
x+1/2=±1/6
x=-1/2±1/6
x=(-3±1)/6
x= -2/3 and -1/3 and of course 3 because of the initial condition...
x=-1/3, -2/3, 3
