First of all, note that #x# can't be zero, otherwise #1/(3x)# would be a division by zero. So, provided #x\ne0#, we can rewrite the equation as
# (3x)/(3x)-8 = 1/(3x) + x (3x)/(3x)#
#iff#
#(-24x)/(3x) = 1/(3x) + (3x^2)/(3x)#
with the advantage that now all the terms have the same denominator, and we can sum the fractions:
#(-24x)/(3x) = (1+3x^2)/(3x)#
Since we assumed #x\ne 0#, we can claim that the two fractions are equal if and only if the numerators are equal: so the equation is equivalent to
#-24x = 1+3x^2#
which leads is to the quadratic equation
#3x^2+24x+1=0#.
To solve this, we can use the classic formula
#\frac{-b\pm sqrt(b^2-4ac)}{2a}#
where #a#, #b# and #c# play the role of #ax^2+bx+c=0#.
So, the solving formula becomes
#\frac{-24\pm sqrt(24^2-4*3*1)}{2*3}#
#=#
#\frac{-24\pm sqrt(576-12)}{6}#
#=#
#\frac{-24\pm sqrt(564)}{6}#
Since #564=36* 47/3#, we can simplfy it out the square root, obtaining
#\frac{-24\pm 6sqrt(47/3)}{6}#
and finally we can simplify the whole expression:
#\frac{-cancel(6)*4\pm cancel(6)sqrt(47/3)}{cancel(6)}#
into
#-4\pm sqrt(47/3)#
