#f(y) = 8y^2-22y+15# is of the form #ay^2+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 #8y^2-22y+15 = (2y-3)(4y-5)#
If #f(y) = 0# then #(2y-3) = 0# giving #y = 3/2#
or #(4y-5) = 0# giving #y = 5/4#.
