1) Let y=2x, then y3-9y2+23y-15=0. Note that if we put y=1 we get:
1-9+23-15=24-24=0, so y=1 is a root of the cubic. We can use synthetic division to divide by this root:
1 | 1 -9 23 -15
1 1 -8 | 15
1 -8 15 | 0 =y2-8y+15=(y-3)(y-5), so y=3 or 5.
x=½y, so, since y=1, 3, 5 x must be ½, 3/2, 5/2.
2) x4+3x2-28=(x2+7)(x2-4)=(x2+7)(x-2)(x+2) has two real roots: 2 and -2 and two complex (imaginary) roots: i√7 and -i√7. (You may not be interested in the complex roots.)