Question 259249
3/x  + 4  =  (x + 1)/x

so 3x=(x+1)*(x+4)

=> 3x = x^2+5x+4
=> x^2+2x+4=0
to solve we can just use the formula for solving a quadratic
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
or follow
=> (x+1)^2 + 3=0
=> x+1= + or - (3)^.5 i  (i - imaginary)
so x= 1+root3 i  and 1-root3 i