We have,
x2+2(a+4)x−5a+64>0
If A>0 and D=B2−4AC<0, then Ax2+Bx+C>0∀x∈R.
Hence,
D<0
⇒4(a+4)2−4(−5a+64)<0
⇒a2+16+8a+5a−64<0
⇒a2+13a−48<0
⇒(a+16)(a−3)<0
⇒a∈(−16,3)
∴ Possible values for a are −15,−14,…….,2 containing 18 integers.
But given range of a is [−5,30], hence a would take values −5,−4,−3,−2,−1,0,1,2, containing 8 integers.
And, [−5,30] has 36 numbers.
∴ Required probability
=368
=92