How do you solve #(2y-3)(2y+3) = 16#?

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Answer 1 · 2016-02-09T05:32:17

Solution is #y=5/2 or −5/2#

#(2y-3) * (2y+3) = 16#

is #4y^2-9=16#

i.e. #4y^2=16+9=25#

i.e. #4y^2-25=0#

Factorizing this it becomes

#(2y-5)*(2y+5)=0#

Hence either #(2y-5)=0# which gives #y=5/2#

or

#(2y+5)=0# which gives #y=-5/2#

Hence solution is #y=5/2 or -5/2#