4+8a2+9a4 has only complex zeroes.
To find them we can use the quadratic formula for a2:
a2=(-8±√(64-144))/18=(-8±√-80)/18=(-4±√-20)/9=(-4±2i√5)/9.
We still have to find a. Start with a2=(-4+2i√5)/9.
a=±√((-4+2i√5)/9)=±⅓√(-4+2i√5). If a=p+iq where p, q are real,
a2=p2+2ipq-q2; equate real and imaginary parts:
p2-q2=-4/9; 2pq=2√5/9, pq=√5/9, q=√5/(9p), q2=5/(81p2);
p2-5/(81p2)=-4/9, 81p4-5=(-4/9)(81p2)=-36p2, 81p4+36p2-5=0=(9p2-1)(9p2+5). Since p is real we can reject the factor 9p2+5.
So 9p2=1, p2=1/9, p=±⅓, q2=p2+4/9=5/9, q=±√5/3; but pq=√5/9 so p and q must have the same sign, that is, (p,q)=(⅓,√5/3) or (-⅓,-√5/3).
Therefore, a=p+iq=±⅓(1+i√5).
We continue with a2=(-4-2i√5)/9. Let a=r+is where r, s are real.
r2-s2=-4/9; 2rs=-2√5/9, rs=-√5/9, s=-√5/(9r), so r=p=±⅓ (see above for p and q).
s2=r2+4/9=5/9, s=-√5/3, because rs<0 so when r>0, s<0 and when r<0, s>0, that is, (r,s)=(⅓,-√5/3) or (-⅓,√5/3).
Therefore, a=r+is=⅓-i√5/3 or -⅓+i√5/3, which can be written ±⅓(1-i√5).
So the 4 zeroes are ⅓(1+i√5), ⅓(1-i√5), -⅓(1+i√5), -⅓(1-i√5).