+27 I'm assuming you're solving \((3x+12)(x+4) = \frac{4x^2+8x+8x^2}{2x}\). With this, start by simplifying the RHS (right hand side) to \(2x+4+4x\), which is equal to \(6x+4\).
Now, expand the LHS (left hand side) to get that \(3x^2+24x+48 = 6x+4\). Shift the stuff on the RHS to the LHS to get \(3x^2+18x+42=0\). Using the quadratic equation, the answers are \(x=-3 + \sqrt{5}i\) and \(x = -3-\sqrt{5}i\) (complex roots).
Hope this helped 
