f(y)=8y2−22y+15 is of the form ay2+by+c with a=8, b=−22 and c=15.
Let A=8, B=22 and C=15.
Look for a factorization of AC=8⋅15=120 into a pair of factors whose sum is B=22.
Notice that A+C=23, which is close to 22, so the pair of factors we are looking for is close to 8,15. Notice that B1=12 and B2=10 works.
Then for each pair (A,B1) and (A,B2), divide by the HCF (highest common factor) to get a pair of coefficients of a factor of f(y) (choosing suitable signs) as follows:
(A,B1)=(8,12)→(2,3)→(2y−3)
(A,B2)=(8,10)→(4,5)→(4y−5)
Hence 8y2−22y+15=(2y−3)(4y−5)
If f(y)=0 then (2y−3)=0 giving y=32
or (4y−5)=0 giving y=54.
