Explanation:

Recalling the formulae ,
(a+b)²=(a−b)²+4ab
Using the above formulae in the given question,
(a+1/a)²=(a−1/a)2+4a.(1/a)
=(a−1/a)2+4
Now, (a−1/a)²+4−2(a−1/a)=12
Let(a−1/a)=z.
=z²−2z+4−12=0
=z²−2z−8=0
=z²−(4−2)z−8=0
= z²−4z+2z−8=0
=z(z−4)+2(z−4)=0
=(z−4)(z+2)=0
Eitherz=4orz=−2
Eithera−1/a=4ora−1/a=−2
Taking first expression,
=a−1/a=4
=(a²−1)/a=4
=a²−4a−1=0
Solving the above quadratic we get
Either a=(4+√(42−4.1.(−1))/2
a=2+√5
Or,a=(4−√(42−4.1(−1))/2
a=2−√5
From second expression,
=(a−1/a)=−2
=(a²−1)/a=−2
=a²−1=−2a
=a²+2a−1=0
Solving the above quadratic we get,
=a=(−2+√(4−4.1.(−1))/2
=a=−1+√2
=a=(−2−√(4−4.1.(−1))/2
a=−1−√2