With questions of this sort, find by trial a value of x for which the equation has a positive result and another for which it has a negative result. You then know that a root must lie in between those x values. If you find a value making the result zero, that's even better, because you found a root straight away. Because this is a cubic equation you only need to find one root, because you can divide by that root to get a quadratic, which is easier to solve. As it happens by trial I find that x=-4 is a root, so I can use synthetic division or algebraic long division to divide the equation by x+4 or the root -4. I get the quadratic to be: 2x^2-6x+5=0. The quadratic has no real roots, because the quadratic formula gives a negative for which there's no square root, so x=-4 is the only real root. However, the quadratic formula gives us complex roots: (6+/-2i)/4=(3+/-i)/2. These complex roots are: x=(3+i)/2 and (3-i)/2, and the real root is -4.