f(x)=2sqrt(3x^5)/3=2x^(5/2)sqrt(3)/3; f'(x)=5x^(3/2)sqrt(3)/3. The equation of the tangent is of the form y=mx+b where m=f'(2)=5*2^(3/2)sqrt(3)/3=5sqrt(8*3)/3. The former expression can be interpreted as either 5sqrt(24)/3=5sqrt(24/9)=5sqrt(8/3)=10sqrt(2/3) or 5sqrt(24/3)=10sqrt(2). So m=5sqrt(8/3) or 10sqrt(2), and when x=2, y=2sqrt(3*32)/3 which is either 2sqrt(96/9)=8sqrt(2/3) or 2sqrt(32)=8sqrt(2).
To find b, put the coords into the equation y=mx+b: 8sqrt(2/3)=20sqrt(2/3)+b, so b=-12sqrt(2/3) or 8sqrt(2)=20sqrt(2)+b, so b=-12sqrt(2). The equation of the tangent at x=2 is y=10xsqrt(2/3)-12sqrt(2/3) or y=10xsqrt(2)-12sqrt(2).