Using sign changes as a guide we see that there are 3 sign changes implying at most 3 positive roots. By altering the signs of the odd-powered terms, there are no sign changes so there are no negative roots. Divide the equation g(x)=0 by the coefficient of x^3, which is 2: x^3-19x^2/2+32x-30. If we have any rational roots, they must be factors of 30, so we have 2, 3, 5, 6, 10, 15. And combining these with the factor we divided by we have 3/2, 5/2, 15/2 as additional candidates. So we start with 3/2: 27/8-171/8+48-30=-144/8+48-30=-18+48-30=0, so 3/2 is a root. Now we divide this root into the original equation using synthetic division:
3/2 | 2 -19..64 -60
........2....3 -24..60
........2 -16..40 | 0
Now we have a quadratic: 2x^2-16x+40=2(x^2-8x+20) which has complex roots. 2(x^2-8x+16+4). If we solve the equation in brackets as (x-4)^2+4=0, x-4=2i so x=4+2i and 4-2i. Therefore the zeroes or roots of g(x) are 3/2, 4+2i, 4-2i. (The question doesn't list any options.)