How do you solve #35=w(w-2) #?

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Answer 1 · 2016-03-19T14:24:32

The solution is #w = 7# and #w = -5#

Firstly change #35 = w(w - 2)# into a standard form
So by factorizing

Expand the bracket

#35 = w^2 - 2w#

then put all terms on the left Hand side (LHS) and equate to zero,
So

#w^2 -2w -35 = 0#

Now factor the quadratic by looking for a pair of factors of #-35# whose sum is #-2#, which are #-7# and #5#:

#(w -7)(w + 5) =0#

For this to be true #w-7=0# or #w+5=0#, so #w=7# or #w=-5#.