y(-1)=2+6+5-18=-5; y(-1.5)=23.625;
y(2.5)=-2.375; y(3)=162-162+45-18=27.
Let zeroes be x=a, x=b, x=c, x=d and -1.5<a<-1, 2.5<b<3.
Let f(x)=y, then f'(x)=df/dx=dy/dx=8x3-18x2+10x.
Newton's Method for n=0, 1, 2, 3, ...
a0=-1;
an+1=an-f(an)/f'(an);
b0=2.5;
bn+1=bn-f(bn)/f'(bn).
We can find approximations for a and b which are increasingly accurate using this method.
We arrive at a=-1.123353 approx and b=2.559903 approx.
We can express these zeroes as a factorised quadratic:
(x-a)(x-b)=(x+1.123353)(x-2.559903).
This expands to the quadratic: x2-x(a+b)+ab=x2-1.436550x-2.875675 approx.
If we divide this quadratic into the original quartic we have:
2x2-3.126899x+6.259401
x2-1.436550x-2.875675 ) 2x4-6.000000x3 +5.000000x2+0.000000x-18.000000
2x4-2.873101x3 -5.751349x2
-3.126899x3+10.751349x2+0.000000x
-3.126899x3 +4.491948x2+8.991945x
6.259401x2-...
We need divide no further because we already have a quadratic quotient.
So the quotient is quadratic 2x2-3.126899x+6.259401 which has no real solutions (but it has two conjugate complex solutions).
Therefore the only real solutions are x=-1.123353 or 2.559903, rounded to 6 decimal places.
To find the complex solutions, you can use the quadratic formula on 2x2-3.126899x+6.259401.
The complicated nature of this solution may indicate that the question contains an error (perhaps a missing x term).